Suppose $X_1, X_2, \ldots$ are jointly continuous and independent, each distributed with marginal pdf $f(x)$, where each $X_i$ is nonnegative. Let $T$ be a random variable defined over the integers $2, 3, \ldots$ and let $\Pr(T=i)$ denote the probability that $X_i > X_1$ and $X_j \leq X_1, \forall j<i$.

a) Write out the explicit expression of $\Pr(T=i)$.

b) Prove that the expectation of $T$ is infinite.

My attempt

\begin{align} \Pr(T=i) &= \int_{x_1\geq x_2,\, \ldots, \,x_1\geq x_{i-1}, \,x_1<x_i} f(x_1,\ldots,x_i)dx_1 \ldots dx_i\\ &=\int_0^\infty\int_0^{x_i}\cdots\int_0^{x_i}\int_{\max(x_1,\ldots,x_{i-1})}^{x_i} f(x_1)\cdots f(x_i)dx_1\ldots dx_i. \end{align}

I'm not sure if this could be further simplified. About the expectation, I don't really have any idea how to proceed. A gentle hint would be appreciated. Many thanks in advance.

p.s. This is actually Exercise 5.2 from Casella & Berger.


It's much simpler than that. $P(T=i)$ is just the probability that $X_i$ is the first value in the sequence which is greater than $X_1$. $P(T=2)$ is trivially $\frac{1}{2}$, since $X_2$ is either greater or smaller than $X_1$ (the "continuous" bit tells us that there are no ties).

See also that the shape of $f(x)$ is not relevant; whatever the sequence, you can assign to each value of the sequence its relative rank. Consider a sequence of $n$. $P(T=n)$ is the probability that the biggest value is $X_n$, the last in the sequence, and the second biggest is $X_1$. So you simply have:

$$ P(T=n) = \frac{1}{n}\frac{1}{n-1} $$

  • $\begingroup$ Wow, I didn't think of it this way, but this solution is totally cool! Thanks! $\endgroup$ – mkmlp Jun 12 at 5:44

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