8
$\begingroup$

Suppose that $Y_1,\dots,Y_{n+1}$ is a random sample from a continuous distribution function $F$. Let$X\sim\mathrm{Uniform}\{1,\dots,n\}$ be independent of the $Y_i$'s. How can I compute $\mathrm{E}\!\left[\sum _{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\right]$?

$\endgroup$
  • 1
    $\begingroup$ This looks like the conditional expectation of a random variable $I$, given that $Y_i \leq Y_{n+1}$. What is $I$? Or, were you trying to write an indicator function? Like $I \{ Y_i \leq Y_{n+1} \}$? $\endgroup$ – GoF_Logistic Mar 8 '17 at 18:13
  • 1
    $\begingroup$ If $I$ is an indicator function just use linearity of expectation. $\endgroup$ – Łukasz Grad Mar 8 '17 at 18:17
  • 2
    $\begingroup$ What do those vertical bars mean in "$I\mid Y_i\le Y_{n+1}\mid$"? This is not a conventional notation for an indicator function, which raises doubts concerning what this question is asking. $\endgroup$ – whuber Mar 8 '17 at 22:39
  • $\begingroup$ Maybe the OP only meant to use the first vertical bar. That could mean that it is intended to mean conditioning. $\endgroup$ – Michael R. Chernick Mar 9 '17 at 2:49
  • 1
    $\begingroup$ @Zen I believe you might have misunderstood: somebody's (your?) edits had fixed the notational problem, not created them! With the rollback, the strange notation has returned. $\endgroup$ – whuber Mar 9 '17 at 16:18
5
$\begingroup$

Here is an alternative answer to @Lucas' using the law of iterated expectations:

$$ \begin{align} E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}\right] & = E\left[E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}|X\right]\right] \\ & = E\left[\sum_{i=1}^XE[1_{(Y_i \leq Y_{n+1})}|X]\right] \\ & = E\left[\sum_{i=1}^XE[1_{(Y_i \leq Y_{n+1})}]\right] \\ & = E\left[\sum_{i=1}^XE\left[E[1_{(Y_i \leq Y_{n+1})}|Y_{n+1}]\right]\right] \\ & = E\left[\sum_{i=1}^XE[F(Y_{n+1})]\right] \\[12pt] & = E[X]E\left[F(Y_{n+1})\right] \\[12pt] & =\frac{n+1}{2}E[F(Y_{n+1})] \end {align}$$

The third step follows from independence of $Y_i$ and $Y_{n+1}$ from $X$; the fourth step is again an application of the law of iterated expectations; the last step is simply an application of the formula for the expectation of a discrete uniform random variable.

By inverting the order of integration, we derive the remaining expectation:

$$ \begin{align} E[F(Y_{n+1})] & = \int_{-\infty}^{\infty}F(y)dF(y) \\ & = \int_{-\infty}^{\infty} \int_{-\infty}^y dF(x)dF(y) \\ & = \int_{-\infty}^{\infty} \int_{x}^{\infty} dF(y)dF(x) \\ & = \int_{-\infty}^{\infty} (1-F(x))dF(x) \\[10pt] & = 1-E[F(Y_{n+1})] \end{align} $$

which implies $E[F(Y_{n+1})] = \frac{1}{2}$. Hence:

$$ E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}\right] = \frac{n+1}{4} $$

$\endgroup$
5
$\begingroup$

By distributional symmetry, $\Pr\{Y_i\leq Y_{n+1}\}=\Pr\{Y_{n+1}\leq Y_i\}$, for each $i=1,\dots,n$. Since $F$ is continuous, we have $$ \Pr\{Y_i\leq Y_{n+1}\} = 1-\Pr\{Y_{n+1}< Y_i\}=1-\Pr\{Y_{n+1}\leq Y_i\}. $$ Therefore, $\mathrm{E}\left[I_{\{Y_i\leq Y_{n+1}\}}\right]=\Pr\{Y_i\leq Y_{n+1}\}=1/2$. Now, we have $$ \mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right] = \mathrm{E}\!\left[\sum_{i=1}^x I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right] = \sum_{i=1}^x\;\mathrm{E}\!\left[I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right] $$ $$ = \sum_{i=1}^x\;\mathrm{E}\!\left[I_{\{Y_i\leq Y_{n+1}\}}\right] = \frac{x}{2}, $$ in which we used the linearity of the conditional expectation and the independence of $X$ and the $Y_i$'s. Hence, $$ \mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\right] = \mathrm{E}\!\left[\mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X\right]\right] = \mathrm{E}\left[\frac{X}{2}\right] = \frac{n+1}{4}. $$

$\endgroup$
3
$\begingroup$

We have \begin{align} E\left[ \sum_{i = 1}^X I[Y_i \leq Y_{n + 1}] \right] &= E\left[ \sum_{i = 1}^n I[i \leq X] I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n E\left[ I[i \leq X] I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n E\left[ I[i \leq X] \right] \cdot E\left[ I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot E[I[Y_i \leq Y_{n + 1}]] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot E\left[ F(Y_{n + 1})] \right] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot \frac{1}{2} \\ &= \frac{n + 1}{4} \end{align}

The second step follows from the linearity of expectations, the third step from the independence of $X$ and $Y_1, ..., Y_{n + 1}$, and the fifth step from the fact that $$F(y) = P(Y \leq y) = E[I[Y \leq y]].$$ To prove the sixth step, you can use partial integration. For the final step, you use the formula for partial sums.

$\endgroup$
  • 1
    $\begingroup$ Where does your $N$ come from? As long as $N>n-1$, that is always going to be true hence you can fix the final result at the value you want. $\endgroup$ – Daneel Olivaw Mar 8 '17 at 20:48
  • 3
    $\begingroup$ If you mean: $N=n$ I do agree with the answer. $\endgroup$ – Daneel Olivaw Mar 8 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.