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579 lines
16 KiB
Org Mode
579 lines
16 KiB
Org Mode
#+TITLE: Probability and Statistics ( BTech CSE )
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#+AUTHOR: Anmol Nawani
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#+LATEX_HEADER: \usepackage{amsmath}
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# *Statistics
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* Ungrouped Data
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Ungrouped data is data that has not been arranged in any way.So it is just a list of observations
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\[ x_1, x_2, x_3, ... x_n \]
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** Mean
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\[ \bar{x} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n} \]
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\[ \bar{x} = \frac{ \sum_{i = 1}^{n} x_i }{n} \]
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** Mode
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The observation which occurs the highest number of time. So the x_i which has the highest count in the observation list.
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** Median
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The median is the middle most observations.
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After ordering the n observations in observation list in either Ascending or Descending order (any order works). The median will be :
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+ n is even
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\[ Median = \frac{ x_\frac{n}{2} + x_{(\frac{n}{2}+1)} }{2} \]
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+ n is odd
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\[ Median = x_\frac{n+1}{2} \]
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** Variance and Standard Deviation
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\[ Variance = \sigma^2 \]
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\[ Standard\ deviation = \sigma \]
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\[ \sigma^2 = \frac{\sum_{i=1}^{n} (x_i - Mean)^2 }{n} \]
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\[ \sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - (Mean)^2 \]
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** Moments
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*** About some constant A
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\[ r^{th}\ moment = \frac{1}{n} \Sigma(x_i - A)^r \]
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*** About Mean (Central Moment)
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When A = Mean, then the moment is called central moment.
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\[ \mu_r = \frac{1}{n} \Sigma(x_i - Mean)^r \]
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*** About Zero (Raw Moment)
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When A = 0, then the moment is called raw moment.
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\[ \mu_r^{'} = \frac{1}{n} \Sigma x_i^r \]
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* Grouped Data
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Data which is grouped based on the frequency at which it occurs. So if 9 appears 5 times in our observations, we group as x(observation) = 9 and f (frequency) = 5.
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#+attr_latex: :align |c|c|c|
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|------------------+---------------|
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| x (observations) | f (frequency) |
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|------------------+---------------|
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| 2 | 5 |
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| 1 | 3 |
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| 4 | 5 |
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| 8 | 9 |
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|------------------+---------------|
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If we store it in data way, i.e. the observations are of form 10-20, 20-30, 30-40 ... then we will get $x_i$ by doing
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\[ x_i = \frac{lower\ limit + upper\ limit}{2} \]
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i.e,
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$x_i$ for 20-30 will be $\frac{20 + 30}{2}$
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So for data
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#+attr_latex: :align |c|c|c|
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|-------+---------------|
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| | f (frequency) |
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|-------+---------------|
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| 0- 20 | 2 |
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| 20-40 | 6 |
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| 40-60 | 1 |
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| 60-80 | 3 |
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|-------+---------------|
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the $x_i$'s will become.
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#+attr_latex: :align |c|c|c|
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|-------+-----+-----|
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| | f_i | x_i |
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|-------+-----+-----|
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| 0- 20 | 2 | 10 |
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| 20-40 | 6 | 30 |
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| 40-60 | 1 | 50 |
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| 60-80 | 3 | 70 |
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|-------+-----+-----|
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** Mean
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\[ \bar{x} = \frac{ \Sigma f_i x_i}{\Sigma f_i } \]
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** Mode
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The *modal class* is the record with the row with the highest f_i
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\[ Mode = l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h \]
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In the formula : \\
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l \rightarrow lower limit of modal class \\
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f_1 \rightarrow frequency(f_i) of the modal class \\
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f_0 \rightarrow frequency of the row preceding modal class \\
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f_2 \rightarrow frequency of the row after the modal class \\
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h \rightarrow size of class interval (upper limit - lower limit)
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** Median
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The median for grouped data is calculated with the help of *cumulative frequency*. The cumulative frequency (cf_i) is given by:
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\[ cf_i = f_1 + f_2 + f_3 + ... + f_i \]
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The *median class* is the class whose cf_i is just greater than or is equal to $\frac{\Sigma f}{2}$
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\[ Median = l + (\frac{(n/2) - cf}{f}) \times h \]
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In the formula : \\
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l \rightarrow lower limit of the median class \\
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h \rightarrow size of class interval (upper limit - lower limit) \\
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n \rightarrow number of observations \\
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cf \rightarrow cumulative frequency of the median class \\
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f \rightarrow frequency of the median class
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** Variance and Standard Deviation
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\[ Variance = \sigma^2 \]
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\[ Standard\ deviation = \sigma \]
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\[ \sigma^2 = \frac{\sum_{i=1}^{n} f_i(x_i - Mean)^2 }{\Sigma f_i} \]
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\[ \sigma^2 = \frac{\sum_{i=1}^n f_ix_i^2}{\Sigma f_i} - (Mean)^2 \]
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** Moments
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*** About some constant A
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\[ r^{th}\ moment = \frac{1}{\Sigma f_i} [\Sigma f_i (x_i - A)^r] \]
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*** About Mean (Central Moment)
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When A = Mean, then the moment is called central moment.
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\[ \mu_r = \frac{1}{\Sigma f_i} [\Sigma f_i (x_i - Mean)^r] \]
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*** About Zero (Raw Moment)
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When A = 0, then the moment is called raw moment.
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\[ \mu_r^{'} = \frac{1}{\Sigma f_i} [\Sigma f_i x_i^r] \]
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* Relation between Mean, Median and Mode
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\[ 3Median = 2Mean + Mode \]
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* Relation between raw and central moments
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\[ \mu_0 = \mu_0^{'} = 1 \]
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\[ \mu_1 = 0 \]
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\[ \mu_2 = \mu_2^{'} - \mu_1^{'2} \]
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\[ \mu_3 = \mu_3^{'} - 3\mu_1^{'}\mu_2^{'} + 2\mu_1^{'3} \]
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\[ \mu_4 = \mu_4^{'} - 4\mu_3^{'}\mu_1^{'} + 6\mu_2^{'}\mu_1^{'2} - 3\mu_1^{'4} \]
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* Skewness and Kurtosis
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** Skewness
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+ If Mean > Mode, then skewness is positive
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+ If Mean = Mode, then skewness is zero (graph is symmetric)
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+ If Mean < Mode, then skewness is zero
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[[./skewness.PNG]]
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*** Pearson's coefficient of skewness
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The pearson's coefficient of skewness is denoted by S_{KP}
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\[ S_{KP} = \frac{Mean - Mode}{Standard\ Deviation} \]
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+ If S_{KP} is zero then distribution is symmetrical
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+ If S_{KP} is positive then distribution is positively skewed
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+ If S_{KP} is negative then distribution is negatively skewed
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*** Moment based coefficient of skewness
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The moment based coefficient of skewness is denoted by \beta_1. The \mu here is central moment.
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\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3} \]
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The drawback of using \beta_1 as a coefficient of skewness is that it *can only tell if distribution is symmetrical or not* ,when $\beta_1 = 0$.
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It can't tell us the direction of skewness, i.e positive or negative.
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+ If \beta_1 is zero, then distribution is symmetrical
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*** Karl Pearson's \gamma_1
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To remove the drawback of the \beta_1 , we can derive Karl Pearson's \gamma_1
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\[ \gamma_1 = \sqrt{\beta_1} \]
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\[ \gamma_1 = \frac{\mu_3}{\mu_2^{3/2}} \]
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+ If \mu_3 is positive, the distribution has positive skewness
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+ If \mu_3 is negative, the distribution has negative skewness
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+ If \mu_3 is zero, the distribution is symmetrical
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** Kurtosis
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Kurtosis is the measure of the peak and the curve and the "fatness" of the curve.
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# https://www.analyticsvidhya.com/blog/2021/05/shape-of-data-skewness-and-kurtosis/
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[[./kurtosis.PNG]]
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# https://www.bogleheads.org/wiki/Excess_kurtosis
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[[./kurtosis2.PNG]]
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The kurtosis is calculated using \beta_2
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\[ \beta_2 = \frac{\mu_4}{\mu_2^2} \]
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The value of \beta_2 tell's us about the type of curve
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+ Leptokurtic (High Peak) when \beta_2 > 3
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+ Mesokurtic (Normal Peak) when \beta_2 = 3
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+ Platykurtic (Low Peak) when \beta_2 < 3
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*** Karl Pearson's \gamma_2
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\gamma_2 is defined as:
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\[ \gamma_2 = \beta_2 - 3 \]
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+ Leptokurtic when \gamma_2 > 0
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+ Mesokurtic when \gamma_2 = 0
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+ Platykurtic when \gamma_2 < 0
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# *Probability
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* Basic Probability
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** Conditional Probability
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If some event B has already occured, then the probability of the event A is:
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\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
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$P(A \mid B)$ is read as A given B. So we are given that B has occured and this is probability of now A occuring.
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** Law of Total Probability
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The law of total probability is used to find probability of some event A that has been partitioned into several different places/parts.
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\[ P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + P(A|B_3)P(B_3) + ... + P(A|B_i)P(B_i) \]
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\[ P(A) = \Sigma P(A|B_i)P(B_i) \]
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*Example*, Suppose we have 2 bags with marbles
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+ Bag 1 : 7 red marbles and 3 green marbles
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+ Bag 2 : 2 red marbles and 8 green marbles
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Now we select one bag at random (i.e, the probability of choosing any of the two bags is equal so 0.5). If we draw a marble, what is the probability that it is a green marble?
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*Sol.* The green marbles are in parts in bag 1 and bag 2. \\
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Let G be the event of green marble. \\
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Let B_1 be the event of choosing the bag 1 \\
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Let B-2 be the event of choosing the bag 2 \\
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Then, $P(G|B_1) = \frac{3}{7 + 3}$ and $P(G|B_2) = \frac{8}{2 + 8}$
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\\
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Now, we can use the law of total probability to get
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\[ P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) \]
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*Example* 2, Suppose a there are 3 forests in a park.
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+ Forest A occupies 50% of land and 20% plants in it are poisonous
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+ Forest B occupies 30% of land and 40% plants in it are poisonous
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+ Forest C occupies 20% of land and 70% plants in it are poisonous
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What is the probability of a random plant from the park being poisonous.
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*Sol.* Since probability is equal across whole area of the park. Event A is plant being from Forest A, Event B is plant being from Forest B and Event C is plant being from Forest C. If event P is plant being poisonous, then using law of total probability,
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\[ P(P) = P(P|A)P(A) + P(P|B)P(B) + P(P|C)P(C) \]
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And we know P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2. Also P(P|A) = 0.20, P(P|B) = 0.40 and P(P|C) = 0.70
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** Some basic identities
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+ Probabilities follow law of inclusion and exclusion
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\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
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+ DeMorgan's Theorem
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\[ P(\overline{A \cap B }) = P(\overline{A} \cup \overline{B}) \]
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\[ P(\overline{A \cup B }) = P(\overline{A} \cap \overline{B}) \]
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+ Some other Identity
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\[ P(\overline{A} \cap B) + P(A \cap B) = P(B) \]
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\[ P(A \cap \overline{B}) + P(A \cap B) = P(A) \]
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* Probability Function
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It is a mathematical function that gives probability of occurance of different possible outcomes. We use variables to represent these possible outcomes called *random variables*. These are represented by capital letters. Example, $X$, $Y$, etc. We use these random variables as:
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\\
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Suppose X is flipping two coins.
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\[ X = \{HH, HT, TT, TH\} \]
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We can represent it as,
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\[ X = \{0, 1, 2, 3\} \]
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Now we can write a probability function $P(X=x)$ for flipping two coins as :
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#+attr_latex: :align |c|c|c|
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|-----+----------|
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| $x$ | $P(X=x)$ |
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|-----+----------|
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| 0 | 0.25 |
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| 1 | 0.25 |
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| 2 | 0.25 |
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| 3 | 0.25 |
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|-----+----------|
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Another example is throwing two dice and our random variable $X$ is sum of those two dice.
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#+attr_latex: :align |c|c|c|
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|-----+----------------|
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| $x$ | $P(X=x)$ |
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|-----+----------------|
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| 2 | $1/36$ |
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| 3 | $2/36$ |
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| 4 | $3/36$ |
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| 5 | $4/36$ |
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| 6 | $5/36$ |
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| 7 | $6/36$ |
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| 8 | $5/36$ |
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| 9 | $4/36$ |
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| 10 | $3/36$ |
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| 11 | $2/36$ |
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| 12 | $1/36$ |
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|-----+----------------|
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** Types of probability functions (Continious and Discrete random variables)
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Based on the range of the Random variables, probability function has two different names.
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+ For discrete random variables it is called Probability Distribution function.
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+ For continious random variables it is called Probability Density function.
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* Proability Mass Function
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If we can get a function such that,
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\[ f(x) = P(X=x) \]
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then $f(x)$ is called a *Probability Mass Function* (PMF).
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** Properties of Probability Mass Function
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Suppose a PMF
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\[ f(x) = P(X=x) \]
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Then,
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*** For discrete variables
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\[ \Sigma f(x) = 1 \]
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\[ E(X^n) = \Sigma x^n f(x) \]
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For $E(X)$, the summation is over all possible values of x.
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\[ Mean = E(X) = \Sigma x f(x) \]
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\[ Variance = E(X^2) - (E(X))^2 = \Sigma x^2 f(x) - ( \Sigma x f(x) )^2 \]
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To get probabilities
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\[ P(a \le X \le b) = \sum_{a}^{b} f(x) \]
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\[ P(a < X \le b) = (\sum_{a}^{b} f(x)) - f(a) \]
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\[ P(a \le X < b) = (\sum_{a}^{b} f(x)) - f(b) \]
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Basically, we just add all $f(x)$ values from range of samples we need.
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*** For continious variables
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\[ \int_{-\infty}^{\infty} f(x) dx = 1 \]
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\[ E(X^n) = \int_{-\infty}^{\infty} x^n f(x) dx \]
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We only consider integral from the possible values of x. Else we assume 0.
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\[ Mean = E(X) = \int_{-\infty}^{\infty} x f(x) dx \]
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\[ Variance = E(X^2) - (E(X))^2 = \int_{-\infty}^{\infty} x^2 f(x) dx - ( \int_{-\infty}^{\infty} x f(x) dx )^2 \]
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To get probability from a to b (inclusive and exclusive doesn't matter in continious).
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\[ P(a < X < b) = \int_{a}^{b} f(x) dx \]
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** Some properties of mean and variance
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+ Mean
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\[ E(aX) = aE(X) \]
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\[ E(a) = a \]
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\[ E(X + Y) = E(X) + E(Y) ]
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+ Variance
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If
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\[ V(X) = E(X^2) - (E(X))^2 \]
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Then
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\[ V(aX) = a^2 V(X) \]
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\[ V(a) = 0 \]
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* Moment Generating Function
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The moment generating function is given by
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\[ M(t) = E(e^{tX}) \]
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** For discrete
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\[ M(t) = \sum_{0}^{\infty} e^{tx} f(x) \]
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** For continious
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\[ M(t) = \int_{-\infty}^{\infty} e^{tx} f(x) dx \]
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** Calculations of Moments (E(X)) using MGF
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\[ E(X^n) = (\frac{d^n}{dt^n} M(t))_{t=0} \]
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* Binomial Distribution
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The use of a binomial distribution is to calculate a known probability repeated n number of times, i.e, doing *n* number of trials.
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A binomial distribution deals with discrete random variables.
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\[ X = \{ 0,1,2, .... n \} \]
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where *n* is the number of trials.
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\[ P(X=x) = \ ^nC_x\ (p)^x(q)^{n-x} \]
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Here
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\[ n \rightarrow number\ of\ trials \]
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\[ x \rightarrow number\ of\ successes \]
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\[ p \rightarrow probability\ of\ success \]
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\[ q \rightarrow probability\ of\ failure \]
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\[ p = 1 - q \]
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+ Mean
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\[ Mean = np \]
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+ Variance
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\[ Variance = npq \]
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+ Moment Generating Function
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\[ M(t) = (q + pe^t)^n \]
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** Additive Property of Binomial Distribution
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For an independent variable $X$. The binomial distribution is represented as
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\[ X ~ B(n,p) \]
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Here,
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\[ n \rightarrow number\ of\ trials \]
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\[ p \rightarrow probability\ of\ success \]
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+ Property
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If given,
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\[ X_1 \sim B(n_1, p) \]
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\[ X_2 \sim B(n_2, p) \]
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Then,
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\[ X_1 + X_2 \sim B(n_1 + n_2, p) \]
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+ *NOTE*
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If
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\[ X_1 \sim B(n_1, p_1) \]
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\[ X_2 \sim B(n_2, p_2) \]
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Then $X_1 + X_2$ is not a binomial distribution.
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** Using a binomial distribution
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We can use binomial distribution to easily calculate probability of multiple trials, if probability of one trial is known. Example, the probability of a duplet (both dice have same number) when two dice are thrown is $\frac{6}{36}$. \\
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Suppose now we want to know the probability of a 3 duplets if a pair of dice is thrown 5 times. So in this case :
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\[ number\ of\ trials\ (n) = 5 \]
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\[ number\ of\ duplets\ we\ want\ probability\ for\ (x) = 3 \]
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\[ probability\ of\ duplet\ (p) = \frac{6}{36} \]
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\[ q = 1 - p = 1 - \frac{6}{36} \]
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So using binomial distribution,
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\[ P(probability\ of\ 3\ duplets) = P(X=3) = \ ^5C_3 \left(\frac{6}{36}\right)^3 \left(\frac{30}{36}\right)^{5-3} \]
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* Poisson Distribution
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A case of the binomial distribution where *n* is indefinitely large and *p* is very small and *$\lambda = np$* is finite.
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\[ P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}\ if\ x = 0, 1, 2 ..... \]
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\[ P(X=x) = 0\ otherwise \]
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\[ \lambda = np \]
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+ Mean
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\[ Mean = \lambda \]
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+ Variance
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\[ Variance = \lambda \]
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+ Moment Generating Funtion
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\[ M(t) = e^{\lambda\left(e^{t}-1\right)} \]
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** Additive property
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If X_1, X_2, X_3..X_n follow poisson distribution with \lambda_1, \lambda_2, \lambda_3....\lambda_n \\
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Then,
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\[ X_1 + X_2 + X_3...+X_n \sim \lambda_1 + \lambda_2 + \lambda_3 + ...+ \lambda_n \]
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* Exponential Distribution
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A continuous random distribution which has probability mass function
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\[ f(x) = \lambda e^{-\lambda x}\ ,\ when\ x \ge 0 \]
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\[ f(x) = 0 \ ,\ otherwise \]
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\[ where\ \lambda > 0 \]
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+ Mean
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\[ Mean = \frac{1}{\lambda} \]
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+ Variance
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\[ Variance = \frac{1}{\lambda^2} \]
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+ Moment Generating Function
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\[ M(t) = \frac{\lambda}{\lambda - t} \]
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** Memory Less Property
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\[ P[X > (s + t) \mid X > t] = P(X > s) \]
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* Normal Distribution
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Suppose for a probability funtion with random variable X, having mean \mu and variance \sigma^2.
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We denote normal distribution using $X \sim N(\mu,\sigma)$ \\
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The probability mass funtion is
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\[ f(x) = \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}\right) \]
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\[ -\infty < x < \infty \]
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\[ -\infty < \mu < \infty \]
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\[ \sigma > 0 \]
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Here, $exp(x) = e^x$
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+ Moment Generating Funtion
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\[ M(t) = exp\left( \mu t + \frac{\sigma^2 t^2}{2} \right) \]
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** Odd Moments
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\[ E(X^{2n + 1}) = 0 \ , \ n = 0, 1, 2, ... \]
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** Even Moments
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\[ E(X^{2n}) = 1.3.5....(2n-3)(2n-1) \sigma^{2n} \ , \ n = 0, 1, 2, ... \]
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** Properties
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+ In a normal distribution
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\[ Mean = Mode = Median \]
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+ For normal distribution, mean deviation about mean is
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\[ \sigma \sqrt{ \frac{2}{\pi} } \]
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** Additive property
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Suppose for distributions X_1, X_2, X_3 ... X_n with means \mu_1 , \mu_2 , \mu_3 ... \mu_n and standard deviation \sigma_1^2 , \sigma_2^2 , \sigma_3^2 ..... \sigma_n^2 respectively.
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\\
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Then X_1 + X_2 + X_3 will have mean *( \mu_1 + \mu_2 + \mu_3 + ... + \mu_n )* and standard deviation *(\sigma_1^2 + \sigma_2^2 + \sigma_3^2 + ..... + \sigma_n^2 )*
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+ Additive Case
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Given,
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\[ X_1 \sim N(\mu_1, \sigma_1) \]
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\[ X_2 \sim N(\mu_2, \sigma_2) \]
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Then,
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\[ a X_1 + b X_2 \sim N \left( a \mu_1 + b \mu_2, \sqrt{ a^2 \sigma_1^2 + b^2 \sigma_2^2} \right) \]
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