Consider $n$ independent trials, each resulting in any one of$r$ possible outcomes with probabilities $P_1,P_2,\dots ,P_r$. Let $X$ denote the number of outcomes that never occur in any of the trials. Find $E\left[X\right]$ and show that among all probability vectors $P_1,\dots ,P_r,E\left[X\right]$ is minimized when$P_i=1/r,i=1,\dots ,r.$

Independent trials $=n$

The$r$possible outcomes with probabilities $P_1,P_2,\dots ,P_r$

the number of outcomes that never occur in any of the trials$=X$

Let's define a new indicator variable as follows:

$X_i=\left\{\begin{array}{c}1\text{if outcome}i\text{did not occur}\\ 0\text{Otherwise}\end{array}\right.$

On a single trial, the probability that event $j$does not occur is given by: $1-P_i$

In $n$trials, the probability that event $i$does not occur is given by: $n·\left(1-P_i\right)$

Now, using the indicator variable defined above, the number of outcomes that never occur in any of the trials is given by:

$X=\sum _i{X}_i$

Hence:

$E\left[X\right]=\sum _{i}E\left[X_i\right]$

$=\sum _{i=1}^{r}n·\left(1-P_i\right)$

$=n·\sum _{i=1}^r\left(1-P_i\right)$

$=n\left[r-\sum _{i=1}^r{P}_i\right]$

The number of outcomes can never be negative. Hence, the expectation value is minimized when it is equal to 0 .

$E\left[X\right]=0$

$n\left[r-\sum _{i=1}^r{P}_i\right]=0$

$r-\sum _{i=1}^r{P}_i=0$

$r=\sum _{i=1}^r{P}_i$

The given equation holds true when:

$P_i=\frac{1}{r}$

Therefore, the value of $E\left[X\right]$is $=n\left[r-\sum _{i=1}^r{P}_i\right]$

The expectation value of$X$is maximized when$P_i=\frac{1}{r}.$