Q.7.15 - A First Course in Probability

Question

Consider $n$ independent trials, the $i th$ of which results in a success with probability $P_l$.

(a) Compute the expected number of successes in the $n$ trials-call it $\mu$

(b) For a fixed value of $\mu$, what choice of $P_1,\dots ,P_n$ maximizes the variance of the number of successes?

(c) What choice minimizes the variance?

Step-by-Step Solution

Answer

a)  The expected number of successes in the $n$ trials-call it $\mu$ is $\sum _{i=1}^n{p}_i=\mu$

b) $Var\left(X\right)=\mu$is the choices of $P_{1...,}{P}_2$maximizes the variance of the number of successes.

c) $Var\left(\mathrm{X}\right)\forall i=1,2,\dots \dots ,n$is the choice minimizes the variance

Step 1: Given Information (Part a)

The $t^{\text{th }}$ independent trial $=n$

Success with probability $=P_i$

The expected number of successes in the $n$ trials $=\mu$

Step 2: Explanation (Part a)

(a) $X=$ Total number of success in $n$ Trials

$=X_1+X_2+\dots \dots +X_n$

Now $\mathrm{E}\left({\mathrm{X}}_{\mathrm{i}}\right)=1.\mathrm{P}\left[{\mathrm{X}}_{\mathrm{i}}=1\right]={\mathrm{p}}_i$

$\Rightarrow \mathrm{E}\left[\mathrm{X}\right]=\sum _{i=1}^n{p}_i=\mu \left(say\right)$

Step 3: Final Answer (Part a)

Hence, the expected number of successes in the $n$ trials-call it $\mu$ is $\sum _{i=1}^n{p}_i=\mu$.

Step 1: Given Information (Part b)

The $t^{\text{th }}$ independent trial $=n$

Success with probability $=P_i$

Fixed value $=\mu$

Step 2: Explanation (Part b)

Calculate the value of $Var\left(X_i\right):$

b)$\begin{array}{r}\mathrm{Var}\left(X_i\right)=E\left[X_i^2\right]-{\left[E\left(X_i\right)\right]}^2\end{array}$

$\begin{array}{r}=p_i-p_i^2\end{array}$

$\begin{array}{r}=p_i\left(1-p_i\right)\end{array}$

$\Rightarrow \mathrm{Var}\left(X\right)=\sum _{i=1}^n \left(p_i-p_i^2\right)$

$=\mu -\sum _{i=1}^n p_i^2$

$\Rightarrow p_i^2\cong ,p_2^2\cong 0,\dots \dots \dots ,p_n^2\cong 0$ maximizes the variance

Maximum value of $Var\left(\mathrm{X}\right)=\mu$

Step 3: Final Answer (Part b)

Therefore, Maximum value of $Var\left(\mathrm{X}\right)=\mu$

Step 1: Given Information (Part c)

The $i^{\text{th }}$ independent trial $=n$

Success with probability $=P_i$

Step 2: Explanation (Part c)

c) $Var\left(X\right)=\mu -\sum _{i=1}^n{p}_i^2$

$Var\left(X\right)=0$is its minimum possible value which is possible if $\mu =\sum _{i=1}^n{p}_i^2$

$\mu =\sum _{i=1}^n p_i^2$

$\Rightarrow p_i=\sqrt{\frac{\mu }{n}} \forall i=1,2,\dots \dots ,n$

$\Rightarrow p_i=\sqrt{\frac{\mu }{n}}$minimizes,

 $Var\left(X\right)\forall i=1,2,\dots \dots ,n$

Step 3: Final Answer (Part c)

Hence,$Var\left(\mathrm{X}\right)\forall i=1,2,\dots \dots ,n$ is the choice minimizes the variance.