Probability that argmax$(X_i + Y_i) = $argmax$X_i$ where $X_i$ and $Y_i$ are standard normal random variables Let $X_i$, $Y_i$ be i.i.d. random variables with standard normal distributions.
How to evaluate
$\mathbb P(\text{argmax}_{1 \le i \le n} (X_i + Y_i) = \text{argmax}_{1 \le i \le n} X_i)$
?
When $n = 2$ the answer should be $3 / 4$. Can one derive a formula for general $n$?
Off-topic: what is the best SE site to ask this question: here, MathOverflow, or Cross Validated?
 A: Let $E_{ik}$ be the event that $X_i\ge X_k$ and $X_i+Y_i\ge X_k+Y_k$. You want $\mathbb P\left(\bigcup_i\bigcap_kE_{ik}\right)$. The events $\bigcap_kE_{ik}$ for different $i$ are disjoint, so we can add their (equal) probabilities, and given $i$, $X_i$ and $Y_i$, the events $E_{ik}$ for $k\ne i$ are independent, so we can multiply their (equal) probabilities. There are $n$ different choices for $i$ and $n-1$ different choices for $k$. Thus, the probability you want is
$$
n\int_{-\infty}^\infty\mathrm dx\,f(x)\int_{-\infty}^\infty\mathrm dy\,f(y)\left(\int_{-\infty}^x\mathrm dx'\,f(x')\int_{-\infty}^{x+y-x'}\mathrm dy'\,f(y')\right)^{n-1}\\
=
n\int_{-\infty}^\infty\mathrm dx\,f(x)\int_{-\infty}^\infty\mathrm dy\,f(y)\left(\int_{-\infty}^x\mathrm dx'\,f(x')F(x+y-x')\right)^{n-1}\;,
$$
where $f$ is the density function of the distribution and $F$ is its cumulative distribution function.
Unfortunately I doubt that these integrals can be solved in analytic form for $n\gt2$ for a normal distribution. For a uniform distribution, however, we get
$$
n\int_0^1\mathrm dx\left[\int_0^{1-x}\left(\frac12x^2+xy\right)^{n-1}\mathrm dy+\int_0^x\left(x-\frac12u^2\right)^{n-1}\mathrm du\right]
$$
(where $u=1-y$), so we just have to integrate polynomials. Here are the results up to $n=7$, together with numerical results for the normal distribution:
$$
\begin{array}{c|c|c|c}
n&\text{uniform}&\text{uniform }\approx&\text{normal }\approx\\\hline
2&\frac34&0.750&0.750\\
3&\frac{77}{120}&0.642&0.644\\
4&\frac{961}{1680}&0.572&0.582\\
5&\frac{1751}{3360}&0.521&0.539\\
6&\frac{21361}{44352}&0.482&0.507\\
7&\frac{2593439}{5765760}&0.450&0.483
\end{array}
$$
Obtaining an asymptotic result for $n\to\infty$ is easier for the uniform distribution. In this case, the maximum will be very close to $1$, and we only need to integrate $P(X_k+Y_k\le1+Y_i)^{n-1}$:
\begin{eqnarray*}
\int_0^1\left(1-\frac12(1-y)^2\right)^{n-1}\mathrm dy
&\approx&
\int_0^1\mathrm e^{-\frac12(n-1)(1-y)^2}\mathrm dy
\\
&=&
\int_0^1\mathrm e^{-\frac12(n-1)u^2}\mathrm du
\\
&\approx&
\int_0^\infty\mathrm e^{-\frac12(n-1)u^2}\mathrm du
\\
&=&
\sqrt{\frac\pi{2(n-1)}}
\\
&\approx&
\sqrt{\frac\pi{2n}}
\;.
\end{eqnarray*}
For the normal distribution, the asymptotic analysis is more involved. I'll try to work it out in more detail when I find the time, but for now, very roughly speaking: The distribution of the maximum has a peak near $\sqrt{\ln n}$, and $f(y)P(X_k+Y_k\le\sqrt{\ln n}+y)^{n-1}$ has a peak near $(\sqrt2-1)\sqrt{\ln n}$, which leads to a leading factor $n^{-(\sqrt2-1)^2}$ in the desired probability. However, there are also factors of order $\sqrt{\ln n}$ that complicate matters.
Here's a log-log plot of numerical results for the probabilities for the uniform (green) and normal (purple) distributions, together with the asymptotic results (red):

The line for the uniform distribution is the result above, $\sqrt{\pi/(2n)}$; the curve for the normal distribution is a linear fit of the logarithm of the probability to $-(\sqrt2-1)^2\ln n+c+d\ln\ln n$, which yields $c\approx-0.131$ and $d\approx-0.359$.
