Q.7.10 - A First Course in Probability

Question

Let $X_1,X_2,\dots ,X_n$ be independent and identically distributed positive random variables. For $k\le n$ find $E\left[\frac{{\sum }_{i=1}^k{X}_i}{{\sum }_{i=1}^n{X}_i}\right].$

Step-by-Step Solution

Answer

The value of $E\left[\frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i}\right]$ is $\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n=k\\ \frac{k}{n},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n>k\end{array}\right..$

Step 1: Given Information

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

Find $E\left[\frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i}\right]=?$

Step 2: Explanation

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

$E\left[\frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i}\right]=\frac{E\sum _{i=1}^k X_i}{E\sum _{i=1}^n X_i}$

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

Step 3: Explanation

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

If $E\left[X_i\right]=\mu$(finite) then:

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

$=\frac{\mu +\mu +\dots \dots +\mu \left(k\text{ times }\right)}{\mu +\mu +\dots \dots +\mu \left(n\text{ times }\right)}$  where $n>k$

$=\frac{k\mu }{n\mu }$

$=\frac{k}{n}$

Therefore:

$E\left[\frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i}\right]=\left\{\begin{array}{l}0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n=k\\ \frac{k}{n},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n>k\end{array}\right.$

Step 4: Final Answer

Hence, the value of $E\left[\frac{{\sum }_{i=1}^k{X}_i}{{\sum }_{i=1}^n{X}_i}\right]$ is $\left\{\begin{array}{l}0,n=k\\ \frac{k}{n},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n>k\end{array}\right..$