Conditional distribution of maximum of i.i.d. normal sample given its sum

Question

I'm trying to find the conditional probability $P(Y_n>a|\bar{X}=x)$ where $Y_n=\max(X_1,...,X_n)$. $X_1,...X_n$ are normal sample from $N(\mu,1)$

My attempt : $$P(Y_n>a|\bar{X}=x)=\frac{P(Y_n>a , \bar{X}=x)}{f_{\bar{X}}(x)}$$

$$P(Y_n>a, \bar{X}=x) = \int_a^\infty f_{Y_n, \bar{X}}(y,x)dy$$ and I know that $n\bar{X}=\sum_1^nY_i$. But I cannot find the joint density of $Y_n$ and $\bar{X}$.

How can I get further from here?

Answer 1

(This is not a full solution but, hopefully, a reduction of the problem to a simpler one.)

Fix $n\geqslant2$ and, for every $1\leqslant k\leqslant n$, introduce $Z_k=X_k-\bar X$. Then $Z=(Z_k)_{1\leqslant k\leqslant n}$ is normal, centered, such that $E(Z_k^2)=1-\frac1n$ and $E(Z_kZ_\ell)=-\frac1n$ for every $k\ne\ell$. Furthermore, $(Z,\bar X)$ is jointly normal and $E(Z\cdot\bar X)=0$, hence $Z$ is independent of $\bar X$.

Let $Y=\max(X_1,\ldots,X_n)$. The preceding remarks show that $Y=W+\bar X$ where $W=\max(Z_1,\ldots,Z_n)$ is independent of $\bar X$. In particular, $$P(Y>y\mid \bar X=x)=P(W>y-x)$$ It remains to compute the distribution of $W$...

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(What follows are some random thoughts on the mathematical setting described above.)

The distribution of $W$ does not depend on $\mu$, and $W$ is the maximum of a gaussian vector of size $n$ and rank $n-1$, namely, the vector $Z=(Z_k)_{1\leqslant k\leqslant n}$, whose distribution is concentrated on the hyperplane $H_n=\{z\in\mathbb R^n\mid \mathbf 1^Tz=0\}$ since $\mathbf 1^TZ=Z_1+\cdots+Z_n=0$ almost surely, or, equivalently, because $\mathbf 1^TC\mathbf 1=0$, where $C=\mathrm{Cov}(Z)$ is the covariance matrix of $Z$, hence $C_{k,k}=1-\frac1n$ and $C_{k,\ell}=-\frac1n$ for every $k\ne\ell$.

Let $(U_k)_{k\geqslant1}$ denote a sequence of i.i.d. standard normal random variables.

• For $n=2$, $W$ is distributed like $$\sqrt{\frac12}|U_1|$$ hence the PDF of $W$ is $$f_W(w)=2\sqrt2\varphi(\sqrt2w)\mathbf 1_{w>0}$$
• For $n=3$, $W$ is distributed like $$\sqrt{\frac2{3}}\max\left(U_1,-\frac12U_1+\frac{\sqrt{3}}2|U_2|\right)$$

(To be continued?)