How to calculate the conditional expectation $E(X_1|Y_n)$?

Question

Let $X_1,X_2, \dots$ be iid $U(0,1)$ and $Y_n$ be the maximum of $X_1, \dots, X_n$. My question is how to calculate $E(X_1 \mid Y_n)$.

My thoughts are as follows, but I don't know whether they're rigorous. \begin{align*} &P(x<X_1<x+\Delta x \mid Y_n=y) \\ =&\lim _{\Delta y \to 0}\dfrac {P(x<X_1<x+\Delta x,y<Y_n<y+Δy)}{P(y<Y_n<y+Δy)} \\ =&\lim _{\Delta y \to 0}\dfrac{ΔxP(y<\max\{X_2,\dots, Xn\}<y+\Delta y)}{P(y<Y_n<y+\Delta y)} \\ =&\dfrac{n-1}{ny}\Delta x(x<y),P(X_1=y \mid Y_n=y)=\dfrac{1}{n}, \end{align*} so under the conditional $Y_n=y$, $X_1$ follows a mixture distribution, we know $$E(X_1 \mid Y_n)=\dfrac{(n+1)Y_n}{2n}.$$ My question is whether this is rigorous? I don't know how to verify this conclusion by measure theory.

Another similar question is whether if $X_1, \dots, X_n$ are iid, then $E(X_1 \mid S_n)=\frac{S_n}{n}$. It seems reasonable by symmetry, but I also can't prove it by measure theory. Any guidance will be greatly appreciated.

Answer 1

For the first part, you are correct. To formalise it, observe that $Y_{n-1} = \max(X_2, \cdots, X_n)$ has a density with respect to the Lebesgue measure on $\mathbb{R}$, $f(y) = (n-1)y^{n-2}\mathbf{1}(y\in(0,1))$. Then fix $A\subseteq \mathbb{R}^n$ Borel measurable and compute directly using independence and Fubini \begin{align} \mathbb{E}[X\cdot \mathbf{1}(Y_n\in A)] &= \displaystyle \int_{[0,1]^2}x\cdot \mathbf{1}(x\lor y \in A)(n-1)y^{n-2}\mathrm{d}x\mathrm{d}y\\ &= \displaystyle \int_{[0,1]^2}x\cdot \mathbf{1}(x\le y)\mathbf{1}(y\in A)(n-1)y^{n-2}\mathrm{d}x\mathrm{d}y\\ & + \displaystyle \int_{[0,1]^2}x\cdot\mathbf{1}(x\ge y)\mathbf{1}(x\in A)(n-1)y^{n-2}\mathrm{d}x\mathrm{d}y\\ & = \frac{n-1}{2n}\mathbb{E}[Y_n \cdot \mathbf{1}(Y_n\in A)]+ \frac{1}{n}\mathbb{E}[Y_n \cdot \mathbf{1}(Y_n\in A)]\\ & = \frac{n+1}{2n}\mathbb{E}[Y_n \cdot \mathbf{1}(Y_n\in A)]\,, \end{align} where I have used again that the law of $Y_n$ has density $f(y) = ny^{n-1}\mathbf{1}(y\in(0,1))$. Since $A$ Borel was arbitrary and this equality characterises conditional expectations, you obtain that almost surely, $$\mathbb{E}[X_1 |Y_n] = \frac{n+1}{2n}Y_n, \quad \mathrm{a.s.}\,.$$

For your second problem, yes, your appeal to symmetry is correct and is essentially a re-labelling argument. To formalise this, you would need the formal definition of conditional expectation and show for any Borel measurable $A\subseteq \mathbb{R}$, \begin{align} \mathbb{E}[X_j \cdot \mathbf{1}(S_n\in A)] &= \displaystyle\int_{[0,1]^n} x_j\mathbf{1}(x_1+\cdots + x_n\in A)\mathrm{d}x_1\cdots \mathrm{d}x_n\\ & \stackrel{\mathrm{Fubini}}{=}\displaystyle\int_{[0,1]^n} x_1\mathbf{1}(x_1+\cdots + x_n\in A)\mathrm{d}x_1\cdots \mathrm{d}x_n\\ & = \mathbb{E}[X_1 \cdot \mathbf{1}(S_n\in A)] = \mathbb{E}[\mathbb{E}[X_1 |S_n] \cdot \mathbf{1}(S_n\in A)]\,, \end{align} by the fact that the random variables are iid and the joint density is unity (so invariant under permutation of indices). Since $A$ Borel was arbitrary and this equality characterises conditional expectations, you obtain that almost surely, $1\le j \le n$ $$\mathbb{E}[X_1 |S_n] = \mathbb{E}[X_j |S_n], \quad \mathrm{a.s.}\,,$$ and the rest should be clear.

Answer 2

See that for $0\leq x< y\leq 1$, $$P(X_{1}\leq x,Y_{n}\leq y)=P(X_{1}\leq x, X_{2}\leq y,...,X_{n}\leq y)=xy^{n-1}$$

So, the joint pdf in this region is given by $f(x,y)=\frac{\partial^{2}}{\partial x \partial y}F(x,y)=(n-1)y^{n-2}\qquad 0\leq x< y\leq 1$

And $Y_{n}$ has pdf given by $f_{Y_{n}}(y)=ny^{n-1}$.

So the conditional pdf in this region is given by $f_{X_{1}|Y_{n}=y}(x)=\frac{n-1}{n\cdot y}\qquad 0\leq x< y$

Additionally, conditional on $Y_{n}=y$, there is a mass of $\frac{1}{n}$ on the singleton $\{y\}$ as $P(X_{1}=y|Y_{n}=y)=P(X_{1}=Y_{n}|Y_{n}=y)=\frac{1}{n}$ as $\sum_{k=1}^{n}P(X_{k}=Y_{n}|Y_{n}=y)=1$ and each probability are equal due to iid-ness.

So the cdf of $X_{1}$ conditional on $Y_{n}=y$ is given by:-

$$F_{X_{1}|Y_{n}=y}(x)=\begin{cases}0\quad ,x<0\\ \frac{n-1}{n}\frac{x}{y}\quad,0\leq x<y\\ 1\quad ,x\geq y \end{cases}$$

, i.e. the cdf has a jump of size $\frac{1}{n}$ at $x=y$.

Thus, \begin{align}E(X_{1}|Y_{n}=y)&=\int_{0}^{\infty}P(X_{1}>x|Y_{n}=y)\,dx=\int_{0}^{\infty}(1-F_{X_{1}|Y_{n}=y}(x))\,dx\\\\ &=\int_{0}^{y}\left(1-\frac{nx}{(n-1)y}\right)\,dx=\frac{(n+1)y}{2n}\end{align}

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Or equivalently

The conditional distribution (measure) of $X_{1}$ on $\Bbb{R}$ given $Y_{n}=y$ is given by $$\mu(A):=\int_{A}\mathbf{1}_{[0,y)}\frac{n-1}{n}\frac{1}{y}\,dx+\frac{1}{n}\delta_{y}(A)$$ for all Borel sets $A\subseteq \Bbb{R}$ where $\delta_{y}$ denotes the Dirac delta measure at $y$.

Thus, \begin{align}E(X_{1}|Y_{n}=y)&=\int_{0}^{y}\frac{n-1}{n}\frac{x}{y}\,dx+\frac{y}{n}\\\\ &=\frac{(n+1)y}{2n}\end{align}

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For the second question, see that by $E(X_{1}|S_{n})=E(X_{j}|S_{n})$ for $1\leq j\leq n$ due to iid distribution.

So, as $E(X_{1}+...+X_{n}|S_{n})=\sum_{k=1}^{n}E(X_{k}|S_{n})=nE(X_{1}|S_{n})$

But the LHS is just $E(S_{n}|S_{n})=S_{n}$. So $E(X_{1}|S_{n})=\frac{S_{n}}{n}$