distribution of one normal RV's rank within another normal distribution This is a question originating from some research I'm doing on the discovery of correlations in a mass of time-series data.
Suppose I have a random variable $X$ that is normally distributed with mean $\mu_X$ and variance $\sigma^2_X$.  In addition, I have i.i.d. random variables $Y_i$ ($1 \leq i \leq n$) that are also normally distributed, but with mean $\mu_Y$ and variance $\sigma^2_Y$.  Typically, $\mu_X > \mu_Y$, and we can assume that's the case in this question.
I'm interested in the distribution of $X$'s rank within the $Y_i$—effectively, the number of $Y_i$ that are greater than $X$.  Is there a straightforward way of obtaining even an approximate expression for this?

There is this question, but I'm not sure that's really sufficiently close to the same question.  For one thing, it concerns only two samples, one drawn from each distribution, and for another, it gives only the probability that $X > Y$.  This is of limited use in determining the distribution of $X$'s rank, since the event $X > Y_i$ is not in general independent of the event $X > Y_j$.  However, if someone can articulate why an answer to that question will give me the answer to mine (or if there is another, more related question and answer), I'd be satisfied with that.
 A: Emm, let me know if I misunderstood anything.
Let's first condition our computations on $X = x.$ 
Since $Y_i$s are normally distributed and i.i.d, we can compute 
$$p(x) = P(x \ge Y_i), Y_i \sim \mathcal{N}(\mu_Y, \sigma_Y^2). $$
The function value of $p(x)$ can be efficiently computed. If you are concerned about the complexity in computing $p(x)$, you can try to search for approximate of the CDF function of normal distributions. I do not remember it on top of my head. But I'v seen some good approximation formulas.
Next, you mentioned that $Y_i$s are i.i.d. Hence
$$ P(X\ ranks\ k, 0\le k \le n \vert X=x) = \binom{n}{k}p(x)^{n-k}(1-p(x))^{k}. $$
Let $f(x)$ be the density function for $\mathcal{N}(\mu_X, \sigma_X^2)$. Then
$$ P(X\ ranks\ k, 0\le k \le n) = \int \left( \binom{n}{k}p(x)^{n-k}(1-p(x))^{k} \right)f(x)dx. $$
To efficiently compute the above formula, especially for large $n$, one can note that the binomial distribution parameterized by $p(x)$ can be approximated by $\mathcal{N}(np(x), np(x)(1-p(x)))$. Refer to this for details/proof.
Let $g(k;x) = \mathcal{N}(np(x), np(x)(1-p(x)))$, you further get
$$ P(X\ ranks\ k, 0\le k \le n) \approx \int g(k;x)f(x)dx. $$
I think most numerical computation packages can handle the above integration now. If desirable, each of the functions $p, f, g$ can be further approximated.
