#+TITLE: Probability and Statistics ( BTech CSE )
#+AUTHOR: Anmol Nawani

# *Statistics

* Ungrouped Data

Ungrouped data is data that has not been arranged in any way.So it is just a list of observations

\[ x_1, x_2, x_3, ... x_n \]

** Mean
\[ \bar{x} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n} \]

\[ \bar{x} = \frac{ \sum_{i = 1}^{n} x_i }{n} \]

** Mode
The observation which occurs the highest number of time. So the x_i which has the highest count in the observation list.

** Median
The median is the middle most observations.
After ordering the n observations in observation list in either Ascending or Descending order (any order works). The median will be :

+ n is even

\[ Median = \frac{ x_\frac{n}{2} + x_{(\frac{n}{2}+1)} }{2} \]

+ n is odd

\[ Median = x_\frac{n+1}{2} \]

** Variance and Standard Deviation

\[ Variance = \sigma^2 \]
\[ Standard\ deviation = \sigma \]

\[ \sigma^2 = \frac{\sum_{i=1}^{n} (x_i - Mean)^2 }{n} \]

\[ \sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - (Mean)^2 \]

** Moments

*** About some constant A

\[ r^{th}\ moment = \frac{1}{n} \Sigma(x_i - A)^r \]

*** About Mean (Central Moment)

When A = Mean, then the moment is called central moment.

\[ \mu_r = \frac{1}{n} \Sigma(x_i - Mean)^r \]

*** About Zero (Raw Moment)

When A = 0, then the moment is called raw moment.

\[ \mu_r^{'} = \frac{1}{n} \Sigma x_i^r \]

* Grouped Data

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.

#+attr_latex: :align |c|c|c|
|------------------+---------------|
| x (observations) | f (frequency) |
|------------------+---------------|
|                2 |             5 |
|                1 |             3 |
|                4 |             5 |
|                8 |             9 |
|------------------+---------------|

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

\[ x_i = \frac{lower\ limit + upper\ limit}{2} \]

i.e, 

$x_i$ for 20-30 will be $\frac{20 + 30}{2}$

So for data

#+attr_latex: :align |c|c|c|
|-------+---------------|
|       | f (frequency) |
|-------+---------------|
| 0- 20 |             2 |
| 20-40 |             6 |
| 40-60 |             1 |
| 60-80 |             3 |
|-------+---------------|

the $x_i$'s will become.

#+attr_latex: :align |c|c|c|
|-------+-----+-----|
|       | f_i | x_i |
|-------+-----+-----|
| 0- 20 |   2 |  10 |
| 20-40 |   6 |  30 |
| 40-60 |   1 |  50 |
| 60-80 |   3 |  70 |
|-------+-----+-----|


** Mean

\[ \bar{x} = \frac{ \Sigma f_i x_i}{\Sigma f_i }  \]

** Mode

The *modal class* is the record with the row with the highest f_i

\[ Mode = l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h  \]

In the formula :  \\
l \rightarrow lower limit of modal class \\
f_1 \rightarrow frequency(f_i) of the modal class \\
f_0 \rightarrow frequency of the row preceding modal class \\
f_2 \rightarrow frequency of the row after the modal class \\
h \rightarrow size of class interval (upper limit - lower limit)

** Median
The median for grouped data is calculated with the help of *cumulative frequency*. The cumulative frequency (cf_i) is given by:

\[ cf_i = f_1 + f_2 + f_3 + ... + f_i \]

The *median class* is the class whose cf_i is just greater than or is equal to $\frac{\Sigma f}{2}$

\[  Median = l + (\frac{(n/2) - cf}{f}) \times h  \]

In the formula : \\
l \rightarrow lower limit of the median class \\
h \rightarrow size of class interval (upper limit - lower limit) \\
n \rightarrow number of observations \\
cf \rightarrow cumulative frequency of the median class \\
f \rightarrow frequency of the median class

** Variance and Standard Deviation

\[ Variance = \sigma^2 \]
\[ Standard\ deviation = \sigma \]

\[ \sigma^2 = \frac{\sum_{i=1}^{n} f_i(x_i - Mean)^2 }{\Sigma f_i} \]

\[ \sigma^2 = \frac{\sum_{i=1}^n f_ix_i^2}{\Sigma f_i} - (Mean)^2 \]

** Moments

*** About some constant A

\[ r^{th}\ moment = \frac{1}{\Sigma f_i} [\Sigma f_i (x_i - A)^r] \]

*** About Mean (Central Moment)

When A = Mean, then the moment is called central moment.
\[ \mu_r = \frac{1}{\Sigma f_i} [\Sigma f_i (x_i - Mean)^r] \]

*** About Zero (Raw Moment) 

When A = 0, then the moment is called raw moment.
\[ \mu_r^{'} = \frac{1}{\Sigma f_i} [\Sigma f_i x_i^r] \]

* Relation between Mean, Median and Mode

\[ 3Median = 2Mean + Mode \]

* Relation between raw and central moments

\[ \mu_0 = \mu_0^{'} = 1 \]
\[ \mu_1 = 0 \]
\[ \mu_2 = \mu_2^{'} - \mu_1^{'2} \]
\[ \mu_3 = \mu_3^{'} - 3\mu_1^{'}\mu_2^{'} + 2\mu_1^{'3} \]
\[ \mu_4 = \mu_4^{'} - 4\mu_3^{'}\mu_1^{'} + 6\mu_2^{'}\mu_1^{'2} - 3\mu_1^{'4} \]

* Skewness and Kurtosis

** Skewness

+ If Mean > Mode, then skewness is positive
+ If Mean = Mode, then skewness is zero (graph is symmetric)
+ If Mean < Mode, then skewness is zero

[[./skewness.PNG]]

*** Pearson's coefficient of skewness

The pearson's coefficient of skewness is denoted by S_{KP}

\[ S_{KP} = \frac{Mean - Mode}{Standard\ Deviation} \]

+ If S_{KP} is zero then distribution is symmetrical
+ If S_{KP} is positive then distribution is positively skewed
+ If S_{KP} is negative then distribution is negatively skewed

*** Moment based coefficient of skewness

The moment based coefficient of skewness is denoted by \beta_1. The \mu here is central moment.

\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3}  \]

The drawback of using \beta_1 as a coefficient of skewness is that it *can only tell if distribution is symmetrical or not* ,when $\beta_1 = 0$.
It can't tell us the direction of skewness, i.e positive or negative.

+ If \beta_1 is zero, then distribution is symmetrical

*** Karl Pearson's \gamma_1

To remove the drawback of the \beta_1 , we can derive Karl Pearson's \gamma_1

\[ \gamma_1 = \sqrt{\beta_1}  \]
\[ \gamma_1 = \frac{\mu_3}{\mu_2^{3/2}}  \]

+ If \mu_3 is positive, the distribution has positive skewness
+ If \mu_3 is negative, the distribution has negative skewness
+ If \mu_3 is zero, the distribution is symmetrical

** Kurtosis

Kurtosis is the measure of the peak and the curve and the "fatness" of the curve. 

# https://www.analyticsvidhya.com/blog/2021/05/shape-of-data-skewness-and-kurtosis/
[[./kurtosis.PNG]]

# https://www.bogleheads.org/wiki/Excess_kurtosis
[[./kurtosis2.PNG]]

The kurtosis is calculated using \beta_2

\[ \beta_2 = \frac{\mu_4}{\mu_2^2} \]

The value of \beta_2 tell's us about the type of curve

+ Leptokurtic (High Peak) when \beta_2 > 3
+ Mesokurtic (Normal Peak) when \beta_2 = 3
+ Platykurtic (Low Peak) when \beta_2 < 3

*** Karl Pearson's \gamma_2

\gamma_2 is defined as:

\[ \gamma_2 = \beta_2 - 3 \]

+ Leptokurtic when \gamma_2 > 0
+ Mesokurtic when \gamma_2 = 0
+ Platykurtic when \gamma_2 < 0

# *Probability

* Basic Probability

** Conditional Probability

If some event B has already occured, then the probability of the event A is:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}  \]

$P(A \mid B)$ is read as A given B. So we are given that B has occured and this is probability of now A occuring.

** Law of Total Probability

The law of total probability is used to find probability of some event A that has been partitioned into several different places/parts.

\[ P(A) = P(A|B_1)P(B_1) +  P(A|B_2)P(B_2) +  P(A|B_3)P(B_3) + ... +  P(A|B_i)P(B_i) \]
\[ P(A) = \Sigma P(A|B_i)P(B_i) \]

*Example*, Suppose we have 2 bags with marbles

+ Bag 1 : 7 red marbles and 3 green marbles
+ Bag 2 : 2 red marbles and 8 green marbles

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?

*Sol.* The green marbles are in parts in bag 1 and bag 2. \\
Let G be the event of green marble. \\
Let B_1 be the event of choosing the bag 1 \\
Let B-2 be the event of choosing the bag 2 \\

Then, $P(G|B_1) = \frac{3}{7 + 3}$ and  $P(G|B_2) = \frac{8}{2 + 8}$
\\
Now, we can use the law of total probability to get 

\[ P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) \]

*Example* 2, Suppose a there are 3 forests in a park. 
+ Forest A occupies 50% of land and 20% plants in it are poisonous
+ Forest B occupies 30% of land and 40% plants in it are poisonous
+ Forest C occupies 20% of land and 70% plants in it are poisonous
What is the probability of a random plant from the park being poisonous.

*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,

\[ P(P) = P(P|A)P(A) + P(P|B)P(B) + P(P|C)P(C) \]

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


** Some basic identities

+ Probabilities follow law of inclusion and exclusion 
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

+ DeMorgan's Theorem
\[ P(\overline{A \cap B }) = P(\overline{A} \cup \overline{B}) \]
\[ P(\overline{A \cup B }) = P(\overline{A} \cap \overline{B}) \]

+ Some other Identity
\[ P(\overline{A} \cap B) + P(A \cap B) = P(B) \]
\[ P(A \cap \overline{B}) + P(A \cap B) = P(A) \]