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Sagot :
The required value [tex]E\left[\frac{\sum_{i=1}^k X_i}{\sum_{i=1}^n X_i}\right][/tex] is [tex]\frac{k_1}{n}[/tex].
As per the details given in the above question are as follow,
let x1,x2,........xn be independent
[tex].$\begin{aligned}& \text { For } k_i \leq n, \\& E {\left[\frac{\sum_{i=1}^{k_1} x_i^2}{\sum_{i=1}^n x_i^2}\right] }\end{aligned}$$\begin{aligned}& E\left[\frac{x_1^2+x_2^2+\cdots+x_n^2}{x_1^2+x_2^2+\cdots+x_n^2}\right]=1 \\& E\left[\frac{x_1^2}{s_n}+\frac{x_2^2}{s_n}+\cdots+\frac{x_n^2}{s_n}\right]=1 \\& \because \quad (x_i^1 \text { are i.i.j) }\end{aligned}$$[/tex]
Further considering,
[tex]\text { Let } \begin{aligned}S_n=x_1^2+x_2^2+\cdots+x_n^2 \\n \in\left[\frac{x_i^2}{S_n}\right]=1\end{aligned}$[/tex]
[tex]E\left[\frac{x_i^2}{S_n}\right] & =\frac{1}{n}, \rightarrow \text { (i) } \\ E\left[\frac{S_k}{S_n}\right] & =E\left[\frac{x_1^2+x_2^2+\cdots x_k^2}{x_1^2+x_2^2+\cdots+x_n^2}\right] \\ & =\left[\frac{x_1^2}{S_n}+\cdots \cdot \frac{x_k^2}{S_n}\right] \\ & =k_1 \cdot E\left[\frac{x_i^2}{S_n}\right] \\ & =k_1 \frac{1}{n} \quad k_1 \leq n \\ & =\frac{k_1}{n} \quad \text { from (i) }[/tex]
To learn more about identical distribution:
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Note: The correct question is,
Let [tex]$X_1, X_2, \ldots, X_n$[/tex] be independent and identically distributed positive random variables. For [tex]$k \leq n$[/tex], find
[tex]E\left[\frac{\sum_{i=1}^k X_i}{\sum_{i=1}^n X_i}\right][/tex]
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