Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $p_1>p_2>...>p_n$.

share|cite|improve this question
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

This is not rigorous, but consider asking for $P(X_i > t)$ for some fixed $t$. If all the $X$'s are normal with mean 0, then we can convert all of them to standard form by dividing by the standard deviations, which are also decreasing. So we can now ask for $P(Z_i > \frac{t}{\sigma_i})$. But $Z_i$ is increasing in $i$ as the numerator is fixed and the denominator is decreasing in $i$. In other words, we have the sequence $Z_1 < Z_2 < Z_3 \ldots$. Therefore, the probability of getting $X_i > t$ is $1 - \Phi(Z_i)$ and this decreases as $Z$ increases. Therefore, it can be shown for some large $t$ that $X_1$ has the largest probability of meeting or exceeding it, so probabilistically it should have the max, and the probability of exceeding a given number is strictly decreasing in the $X_i$'s as long as the variance is strictly increasing.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.