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Let $X_1, ..., X_n$ be independent random variables with the same distribution such that $E(X_1^{-1})$ exists. Show that for $m \leq n$ the expected value

$E(\frac{S_m}{S_n}) = \frac{m}{n}$,

where $S_n = X_1 + ... + X_n$.

Writing this directly, I have:

$E(\frac{S_m}{S_n}) = E(\frac{X_1 + ... + X_m}{X_1 + ... + X_m + ... + X_n})$

I know that the expected value of a sum of independent random variables is the sum of their expected values and I understand that since they have the same distribution, their expected values are the same. However, I don't know how to deal with such fraction, any suggestions?

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  • $\begingroup$ For $X_1^{-1}$, do you mean $1/X_1$? $\endgroup$ – Jack Jan 10 '17 at 0:19
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    $\begingroup$ apply the linearity to the numerator, then $\mathbb{E}\frac{S_m}{S_n}=m\mathbb{E}\frac{X_1}{S_n}$, and finish by symmetry. $\endgroup$ – Ákos Somogyi Jan 10 '17 at 0:19
  • $\begingroup$ @Jack I thinks so, that's the way it's written in the notes and it's not explained $\endgroup$ – Angie Jan 10 '17 at 0:20
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    $\begingroup$ @Angie I think $E[\frac{1}{X_1}]$ existing means that $X_1=0$ can never happen $\endgroup$ – Noble Mushtak Jan 10 '17 at 0:21
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    $\begingroup$ Not that, but use that iid implies $\mathbb{E}\frac{X_i}{S_n}=\mathbb{E}\frac{X_j}{S_n}$ if $i,j\leq n$. Then summing these together: $$1=\sum_{i=1}^n\mathbb{E}\frac{X_i}{S_n}$$ $\endgroup$ – Ákos Somogyi Jan 10 '17 at 1:05
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By linearity of expectation you know that

$$ \mathbb{E}[S_m/S_n] = \sum_{i=1}^m \mathbb{E}[X_i/S_n] = m \mathbb{E}[X_1/S_n]$$ because the $X_i$ are i.i.d. Finally note that $$ \mathbb{E}[S_n/S_n] = \mathbb{E}[1] = 1 = \sum_{i=1}^n \mathbb{E}[X_i/S_n]$$ and again since the $X_i$ are i.i.d., we notice that $$ \sum_{i=1}^n \mathbb{E}[X_i/S_n] = n \mathbb{E}[X_1/S_n] = 1 \Longrightarrow \mathbb{E}[X_1/S_n] = \frac1n$$ Combining these results, you get your claim.

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