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$\{X_i\}$ are i.i.d. standard Normal variables, with mean 0 and standard deviation 1, $$S={1\over{n}}\sum_{i=1}^nX_i$$ What is the conditional probability $P(X_1|S\ge1)$?

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up vote 3 down vote accepted

Call $\varphi$ the standard normal PDF and $\Phi$ its CDF. For each $n\geqslant2$, the density $g_n$ of the desired conditional distribution is such that $$ g_n(x)=\frac{\varphi(x)\,\Phi\left(\frac{x-n}{\sqrt{n-1}}\right)}{\Phi(-\sqrt{n})}. $$

To see this, fix a bounded measurable function $u$ and compute $$ E(u(X_1); S\geqslant 1)=E(u(X_1);X_1+\sqrt{n-1}Y\geqslant n), $$ where $Y=\frac1{\sqrt{n-1}}(X_2+\cdots+X_n)$ is standard normal and independent of $X_1$. Thus, $$ E(u(X_1); S\geqslant 1)=\iint u(x)\mathbf 1_{x+\sqrt{n-1}y\geqslant n}\varphi(x)\varphi(y)\mathrm dx\mathrm dy. $$ Note that, for every fixed $x$, $$ \int\mathbf 1_{x+\sqrt{n-1}y\geqslant n}\varphi(y)\mathrm dy=1-\Phi\left(\frac{n-x}{\sqrt{n-1}}\right)=\Phi\left(\frac{x-n}{\sqrt{n-1}}\right), $$ hence $$ E(u(X_1)\mid S\geqslant 1)=\frac1{P(S\geqslant1)}\int u(x)\varphi(x)\Phi\left(\frac{x-n}{\sqrt{n-1}}\right)\mathrm dx. $$ Finally, $S$ is centered normal with variance $n$ hence $P(S\geqslant1)=1-\Phi(\sqrt{n})=\Phi(-\sqrt{n})$, and the expression of $E(u(X_1)\mid S\geqslant 1)$ computed above, holds for every bounded measurable function $u$, hence the density $g_n$ follows.

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\begin{equation} \begin{split} P(X_1\leq x|S\geq 1)=&\int_{-\infty}^{x}P(S\geq 1|X_1=u)p_{X_1}(u)du\\ \ =&\int_{-\infty}^{x}P(1/N\sum_{i\ne 1}X_i\geq 1-X_1/N|X_1=u)p_{X_1}(u)du\\ \end{split} \end{equation} Now $1/N\sum_{i\ne 1}X_i \perp X_1$ and $1/N\sum_{i\ne 1}X_i\sim \mathcal{N}(0,1/(N-1))$ So we get required probability =\begin{split} \int_{-\infty}^{x}Q\left(\frac{1-u/N}{\sqrt{\frac{1}{N-1}}}\right)p_{X_1}(u)du\\ \end{split} Where $p_{X_1}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2\pi}} $ I think there must be a closed form expression to this integral, but I'm too tired to derive it now.


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As $N$ goes to infinity, this conditional probability does not make sense, because the probability of the event $\hat{S} \geq 1$ goes to 0 by the LLN.

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This is not true. – Did Jul 14 '13 at 9:08

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