Let $X_1,X_2,...,X_n$ be i.i.d uniform random variables on $[0,1]$. Define $Y=max${$X_1,X_2,...,X_n$}. Find $E[X_1|Y]$.

My answer: $P(X_i|Y)=P(X_j|Y)$ for all $i,j=1,2,...,n$.
Since $\sum_{i=n}^{n}P(X_i|Y)=1$, I get $P(X_i|Y)=1/n$.
Then $$E[X_1|Y]=\int_{0}^{1}x_1\times{1\over n}dx_1={1\over 2n}$$ Is it correct for $P(X_i|Y)=P(X_j|Y)$ and $\sum_{i=n}^{n}P(X_i|Y)=1$?


The equality $\sum_i \mathbb P[X_i|Y]=1$ is false, you don't get the right result for this reason.

Observe that $\mathbb P[Y\leq y] = \mathbb P[X_1\leq y\land X_2\leq y\land\dots\land X_n\leq y]=y^n$ hence $f_Y(y)=n y^{n-1}$. Similarly $\mathbb P[X_1\leq x,Y\leq y]=\mathbb P[X_1\leq \min(x,y)\land X_2\leq y\land\dots\land X_n\leq y]=\min(x,y) y^{n-1}$, hence for $x\leq y$, $f_{X_1,Y}(x,y)=(n-1) y^{n-2}$.

Using that, you can obtain the conditional distribution of $X_1$ given that $Y$, for $x<y$ \begin{align*} f_{X_1|Y}(x|y) &= \frac{f_{X_1,Y}(x,y)}{f_Y(y)}\\ &=\frac{(n-1)y^{n-2}}{n y^{n-1}}\\ &=\frac{n-1}{ny} \end{align*}

For $x=y$ a discontinuity happens and $\mathbb P[X_1=y|Y=y]=\frac{1}{n}$ by symmetry. Hence for any $x\leq y$, \begin{align*} f_{X_1|Y}(x|y) =\frac{n-1}{ny}+\frac{1}{n}\delta(x-y) \end{align*}

finally the conditional expectation is given by \begin{align*} \mathbb E[X_1|Y=y] &= \int_0^{y} x \cdot f_{X_1|Y}(x|y) dx\\ &=\frac{n-1}{ny} \cdot \frac{y^2}{2}+\frac{y}{n}\\ &=\frac{y(n+1)}{2n} \end{align*}

And hence $\mathbb E[X_1|Y]=\frac{Y(n+1)}{2n}$

  • 1
    $\begingroup$ But you ignored the point mass with weight $1/n$ of $f_{X_1 \mid Y}(x \mid y)$ at $x = y$. If $f_{X_1 \mid Y} (x \mid y) = (n-1)/(ny)$ then when you integrate $\int_0^y f_{X_1 \mid Y} (x \mid y) dx$ you get $(n - 1)/n$, not $1$. By the symmetry that the OP was possibly referring to, we have $P(X_1 = Y) = 1/n$. $\endgroup$ – Alex Jan 12 '19 at 22:42
  • $\begingroup$ @Alex Oh yes, something felt wrong, thank you very much. $\endgroup$ – P. Quinton Jan 12 '19 at 22:50
  • $\begingroup$ @P.Quinton what does the $\delta$ in case $x=y$ mean? $\endgroup$ – clement Jan 13 '19 at 2:12
  • $\begingroup$ @P.Quinton and why don't you consider $x\gt y$? For $x\ge y$, $f_{X_1,Y}(x,y)=ny^{n-1}$, thus $E[X_1|Y]=1/2$ $\endgroup$ – clement Jan 13 '19 at 5:39
  • $\begingroup$ The Dirac delta : en.wikipedia.org/wiki/Dirac_delta_function is a handy trick to represent some of the mixed random variables that should not have a pdf. It's properties are $\int_A f(x) \delta(x) dx = f(0)$. In this case it means that if you integrate $\int_A f(x) \delta(x-y) dx$ on a range $A$ that contains $y$, then it is equal to $f(y)$. This is a informal trick since $\delta$ is not really a function. To make it formal, you can define it as a measure (the approach taken in probability), however this is digging in the time consuming world of measure theory. $\endgroup$ – P. Quinton Jan 13 '19 at 7:33

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