91Ó°ÊÓ

Independent and identically distributed positive random variables are $X_1,X_2,…,X_n$.

Find$E\left[\frac{\underset{i=1}{\overset{k}{∑}} X_i}{\underset{i=1}{\overset{n}{∑}} X_i}\right]=?$

It is formally perceived that the sum of the independent and identically distributed random variables approaches a normal distribution with the mean $\mathrm{μ}$and the standard deviation $\mathrm{σ}$. As given in the question:

$E\left[\frac{\underset{i=1}{\overset{k}{∑}} X_i}{\underset{i=1}{\overset{n}{∑}} X_i}\right]=\frac{E\underset{i=1}{\overset{k}{∑}} X_i}{E\underset{i=1}{\overset{n}{∑}} X_i}$

$=\frac{E\left[X_1+X_2+……+X_k\right]}{E\left[X_1+X_2+……+X_n\right]}$

The expectation of sums of random variables is always equals to the sum of the expectation of each random variable.

If $E\left[X_i\right]=\mathrm{μ}$(finite)then:

$\frac{E\left[X_1+X_2+……+X_k\right]}{E\left[X_1+X_2+……+X_n\right]}=\frac{E\left[X_1\right]+E\left[X_2\right]+……+E\left[X_k\right]}{E\left[X_1\right]+E\left[X_2\right]+……+E\left[X_n\right]}$

$=\frac{\mathrm{μ}+\mathrm{μ}+……+\mathrm{μ}\left(k\text{times}\right)}{\mathrm{μ}+\mathrm{μ}+……+\mathrm{μ}\left(n\text{times}\right)}$ where$n>k$

$=\frac{k\mathrm{μ}}{n\mathrm{μ}}$

$=\frac{k}{n}$

Therefore:

$E\left[\frac{\underset{i=1}{\overset{k}{∑}} X_i}{\underset{i=1}{\overset{n}{∑}} X_i}\right]=\left\{\begin{array}{l}0,â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots n=k\\ \frac{k}{n},â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots n>k\end{array}\right.$

Hence, the value of $E\left[\frac{{∑}_{i=1}^k{X}_i}{{∑}_{i=1}^n{X}_i}\right]$is$\left\{\begin{array}{l}0,n=k\\ \frac{k}{n},â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots n>k\end{array}\right..$