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

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

It is formally perceived that the sum of the independent and identically distributed random variables approaches a normal distribution with the mean $渭$and the standard deviation $蟽$. As given in the question:

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

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

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

If $E\left[X_i\right]=渭$(finite)then:

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

$=\frac{渭+渭+鈥�鈥�+渭\left(k\text{times}\right)}{渭+渭+鈥�鈥�+渭\left(n\text{times}\right)}$ where$n>k$

$=\frac{k渭}{n渭}$

$=\frac{k}{n}$

Therefore:

$E\left[\frac{\underset{i=1}{\overset{k}{鈭�}}鈥�X_i}{\underset{i=1}{\overset{n}{鈭�}}鈥�X_i}\right]=\left\{\begin{array}{l}0,鈥呪赌呪赌呪赌�n=k\\ \frac{k}{n},鈥呪赌呪赌呪赌�n>k\end{array}\right.$

Hence, the value of $E\left[\frac{{鈭�}_{i=1}^k{X}_i}{{鈭�}_{i=1}^n{X}_i}\right]$is$\left\{\begin{array}{l}0,n=k\\ \frac{k}{n},鈥呪赌呪赌呪赌�n>k\end{array}\right..$