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Let $X_1$, $X_2$, ..., $X_n$ be independent normal variables with zero mean and decreasing variances: $X_i=N(0,v_i)$ where $v_1>v_2>...>v_n\ge 0$. Assume $n>2$. Let $p_i=Prob(X_i = max(X_1,X_2,...,X_n))$.

Prove the following
Conjecture: $p_1>p_2>...>p_n$.

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Interesting question, but you should indicate why you think it is true and what you tried to solve it. – Did Dec 2 '12 at 23:50
Having $n$ random variables is a distraction. You may try to prove your assertion conditioned on $n-2$ variables. That is, try to show that when $i>j$, $Pr(X_i>\max(X_j,t))$ is larger than $Pr(X_j>\max(X_i,t))$ for any fixed $t$. Then you are done. – user1551 Dec 2 '12 at 23:57
Some intuition may be obtained by considering a simpler problem, where $X_i = s_i Y_i$, where $s_i = sqrt(v_i)$, and where $Y_i$ is +1/-1 with equal probability. Then $X_1$ is the max with probability 1/2, $X_2$ the max with probability 1/4, and so on. – user51534 Dec 3 '12 at 1:42
I submitted & deleted this erroneous answer, but am editing out the erroneous part for what its worth 2. $p_i = \int \frac {\phi_i(x)} {\Phi_i(x)} \prod \limits_{j=1}^n \Phi_j(x))$ which is of the form $ \int \frac {\phi_i(x)} {\Phi_i(x)} h(x), h \uparrow$. Fix all parameters that occur in $h$ and look at this as $p(t) = \int \frac {t\phi(t x)} {\Phi(t x)} h(x)$. Then $p_i = p(\frac 1 {\sigma_i})$. $\frac {t\phi(t x)} {\Phi(t x)} = \frac d {dx} $ so $p(t) = -\int log \Phi(t x) h^{\prime} $, $p^{\prime}(t) = -\int (\frac d {dt} {log \Phi(t x)} ) h^{\prime}(x)$ – mike Dec 4 '12 at 13:32

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