Question

Consider $n$independent trials, the $ith$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?

Short 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-by-step solution

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 $=\mu$

Explanation (Part a) 

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

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

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

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

Final Answer (Part a) 

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

Given Information (Part b) 

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

Success with probability$=P_i$

Fixed value$=\mu$

Explanation (Part b)  

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

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

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

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

role="math" localid="1647525060894" $\Rightarrow \mathrm{Var}\left(X\right)=\sum _{i=1}^n \left(p_i-p_i^2\right)$

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

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

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

Final Answer (Part b)  

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

Given Information (Part c) 

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

Success with probability$=P_i$

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,

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

Final Answer (Part c)  

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