# Expected number of samples above certain value of a normally distributed variable with a given sample mean

Suppose $n$ values, $X_1,...,X_n,$ are generated by a random number generator with normal distribution $N(0,1).$ Suppose that the (sample) mean of $X_1,...,X_n$ is $\mu.$ What is known about the order statistics of $X_1,...,X_n$? (Eg. what is the expected value of $Y_c=\#\{i: X_i>c\}$ as an expression in $n,\mu$ and $c.$ If there is no exact formula, then perhaps value of limit of it as $n\to \infty$?)

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I would recommend you include some context and motivation in your question. The way you present it, it somehow resembles an excercise or a homework problem. This (mis)conception could explain the poor reception. –  quid Jun 7 at 11:25
Thanks. I edited the question a bit. It is definitely not a hw question. Since $E(Y_c)$ is not continuous function of $\mu,c$ for $n=1$, I am not sure if there is a simple formula for it for $n>1.$ –  Adam S Sikora Jun 7 at 13:54

Hint: the conditional distribution of $X_i$ given $X_1 + \ldots + X_n = \mu$ is normal ...
$X_1$ and $S = X_1 + \ldots + X_n$ are jointly normal with means $0$ and covariance matrix $C = \pmatrix{1 & 1 \cr 1 & n\cr}$. So they have joint density $$f_{X_1,S}(x,s) = \dfrac{1}{2\pi \sqrt{n-1}} \exp\left(-\dfrac{n x^2 - x s + s^2}{2(n-1)}\right)$$ From that you can compute that the conditional distribution of $X_1$ given $S=s$ is normal with mean $s/n$ and variance $1 - 1/n$. In particular, $$P(X_1 > c| S = n \mu) = 1 - \Phi\left(\dfrac{c-\mu}{\sqrt{1-1/n}}\right)$$ where $\Phi$ is the standard normal CDF. The same is true for all other $X_i$, so by linearity of expected value
$$E[Y_c|S = n \mu] = n \left(1 - \Phi\left(\dfrac{c-\mu}{\sqrt{1-1/n}}\right)\right)$$