We can consider an alterative formula for evaluation of the variance.

$\operatorname{Var}(X)=E(X-\mu)^2 \quad\\ \text { where } \mu=E(X)$

$\operatorname{Var}(X)=E\left[X^2-2 \mu X+\mu^2\right] \quad \ldots(1)$

To simplify (1), we will use some properties of expectation.

$\text { (i) }$

$E(c) =c$

$E(c) =c \cdot p_1+c \cdot p_2+\cdots+c \cdot p_n$

$=c(p_1+\cdots+p_n)=c\cdot 1=c$

Score of the card problem

$E\left(x^2\right)=\sum_{i=2}^{10} i^2 \cdot \frac{1}{13}+(15)^2 \cdot \frac{3}{13}+(18)^2 \cdot \frac{1}{13}$

$=\frac{1}{13}\left(\frac{10 \times 11 \times 21}{6}-1\right)+\frac{225\times3}{13}+\frac{324}{13} =\frac{1383}{13}$

$\operatorname{Var}(X)={E(X^2)}-\{E(X)\}^2 =\frac{1383}{13}-(9)^2=\frac{330}{13}$

Standard deviation of a random variable $X$ is defined as

$\text { s.d. (X) }=\sqrt{\operatorname{Var}(X)}$

Two distinct fair dies are rolled once.

Let $X$ denote the absolute difference of the numbers shown on the upper faces of the two dice.

Find the distribution of $X$ and Var(X).

• Solution:

The possible values of $X$ (absolute difference) are $0,1,2,3,4,5$.

The absolute difference is $0$ for

$\{ (1,1), (2,2), (3,3),(4,4),(5,5),(6,6)\}$

So, $P(X=0)=\frac{6}{36}=\frac{1}{6}$

The absolute difference is $1$ for

$\{(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4)(5,6),(6,5)\}$ $~$ So, $P(X=1)=\frac{10}{36}=\frac{5}{18}$

So the probability distribution of $x$ is

$p_0=\frac{1}{6}, p_1=\frac{5}{18}, p_2=\frac{2}{9}, p_3=\frac{1}{6}, p_4=\frac{1}{9}, p_5=\frac{1}{18}$

$E(X)=0 \cdot \frac{1}{6}+1 \cdot \frac{5}{18}+2 \cdot \frac{2}{9}+3 \cdot \frac{1}{6}+4 \cdot \frac{1}{9}+5 \cdot \frac{1}{18}$ $~$ $=\frac{35}{18}$

$E\left(X^2\right)=0^2 \cdot \frac{1}{6}+1^2 \cdot \frac{5}{18}+2^2 \cdot \frac{2}{9}+3^2 \cdot \frac{1}{6}+4^2 \cdot \frac{1}{9}+5^2 \cdot \frac{1}{18}$

$=\frac{35}{6} \text {. }$ $~$ $\operatorname{Var}(X)=E\left(X^2\right)-[{E(X)}]^2=\frac{35}{6}-\left(\frac{35}{18}\right)^2=\frac{665}{324} \text {. }$

$\underline{\text{Example}}$: An urn contains $n$ distinct marbles. Marbles are drown successively with replacement from the urn at random until a marble is repeated.

Let $X$ denote the number of draws needed to complete the experiment. Find the distribution of $X$.

Solution: $X$ can take values $2,3, \cdots,(n+1)$ $P(X=2)=\frac{1}{n}$ (ie. the marble drawn on the first draw is drawn again)

$P(X=3)=\frac{n-1}{n} \cdot \frac{2}{n}$

(ie. the marble on second is distinct from the first and third is any of the first row)

In many real life experiments, one is interested in only two possible outcomes.

For example: hitting a target

$\quad \quad \downarrow$

(success or failure)

$\rightarrow$ treating a patient

$\quad \quad \downarrow$

(cured or not cured)

$\rightarrow$ appearing in exam

$\quad \quad \downarrow$

(qualifying not qualifying)

In such experiments, we specify, one outcome as success (s) another outcome as failure (f)

These are called Bernoullian trials.

Suppose $n$ Bernoullian trials are performed under identical conditions & independently of each other. Suppose the probability of success is same as all trials (say $p$ ) & so probability of failure is also the same in all trials.

(say$\ (1-p)=q),\\0<p<1 \quad \text{and}\quad 0<q<1$

Let $X \rightarrow$ the number of successes in $n$ trials.

Then $X$ is a discrete $r . v$. and it can take values $0,1,2, \ldots, n$

$p_k=p(x=k)\\= \binom{n}{k} p^k q^{n-k}, \quad k=0,1,2, \ldots, n$

This is called binomial distribution.

We first show that this is a valid probability distribution.

$\text { i.e. } \sum_ {k=0}^n p_ k ~ =\sum_{k=0}^n\binom{n}{k} p^k q^{n-k}$

$= \binom{n}{0} q^n+ \binom{n}{1} p q^{n-1}+\binom{n}{2} p^2 q^{n-2}+\cdots+\binom{n}{n} p^n$ $= (q+p)^n=(1)^n =1$

As $0<q<1, \quad n p q<n p$

So in a binomial distribution, the mean is always more than the variance.

The standard deviation

$=$ s.d. $(X)=\sqrt{n p q}$.

Special case: when the success & failure probabilities are the same,

i.e. $p=q=\frac{1}{2}$,

then the distribution $^n$ takes a very simple form

$P(X=k)=\binom{n}{k}\cdot \frac{1}{2^n}, \quad k=0,1 \ldots, n$

In this case,

$P(X=n-k) = \binom{n}{n-k} \frac{1}{2^n}$

$= \binom{n}{n-k} \cdot \frac{1}{2^n}=P(X=k)$

For example

$p_0=p_n=\frac{1}{2^n}$

$p_1=p_{n-1}=\frac{n}{2^n} \ldots$

In this case,

$E(X)=\frac{n}{2},\\V(X)=\frac{n}{4}$

$\text { s.d. }(X)=\frac{\sqrt{n}}{2} \text {. }$