$\newcommand{\E}[1]{{\textbf E}\left[ #1 \right]} \newcommand{\Var}[1]{{\textbf{Var}}\left[ #1 \right]} \renewcommand{\Pr}[1]{{\textbf{Pr}}\left( #1 \right)}$

I'm trying to find the expected value for the maximum of a bunch of nonegative random variables (geometrically distributed). But when I broke apart an infinite sum and shifted the index of one part to make terms cancel out, I got a negative result for the expected value.

There are $n$ independently and identically distributed geometrically random variables, $X_1... X_n$. I was trying to bound the expected value of the maximum of these random variables, $X = \max_i X_i$. Each of the $X_i$ can take integer values between $0$ and $\infty$. (and the geometric distribution isn't really part of the problem. I think this can happen with any distribution that has an integer support that goes to infinity)

Here's the faulty reasoning:

Since the $X_i$ are independent, for any $j$ in the support of the $X_i$ $$\Pr{X \le j} = \prod_i \Pr{X_i \le j}$$.

Also, $$\Pr{X=j} = \prod_{i} \Pr{X_i\le j} - \prod_{i}\Pr{X_i\le j-1}$$.

So \begin{align*} \E{X} &= \sum_{j=0}^\infty (j)\Pr{X=j} \\ &= \sum_{j=0}^\infty (j)\left(\prod_{i} \Pr{X_i\le j} - \prod_{i}\Pr{X_i\le j-1}\right)\\ &= \sum_{j=0}^\infty (j)\left(\prod_{i} \Pr{X_i\le j}\right) - \sum_{j=0}^\infty(j)\left(\prod_{i}\Pr{X_i\le j-1}\right)\\ &= \sum_{j=0}^\infty (j)\left(\prod_{i} \Pr{X_i\le j}\right) - \sum_{j=1}^\infty(j)\left(\prod_{i}\Pr{X_i\le j-1}\right) & \Pr{X_i \le -1} = 0\\ &= \sum_{j=1}^\infty (j)\left(\prod_{i} \Pr{X_i\le j}\right) - \sum_{j=1}^\infty(j)\left(\prod_{i}\Pr{X_i\le j-1}\right) & 0\cdot\Pr{X_i \le 0} = 0\\ &= \sum_{j=1}^\infty (j)\left(\prod_{i} \Pr{X_i\le j}\right) - \sum_{j=0}^\infty(j+1)\left(\prod_{i}\Pr{X_i\le j}\right) & \text{index shift}\\ &= \sum_{j=1}^\infty \left(j-(j+1)\right) \left(\prod_i(\Pr{X_i\le j}\right) - \prod_i \Pr{X_i \le 0} & \text{rearrangement}\\ &= \sum_{j=1}^\infty \left(-1)\right) \left(\prod_i(\Pr{X_i\le j}\right) - \prod_i \Pr{X_i \le 0} \end{align*}

Which should always negative! Why is this happening? Was it because I broke apart an infinite sum?

  • 1
    $\begingroup$ Your suspect should be correct. Several things to note: You are trying to derive the formula $\displaystyle E[X] = \sum_{j=0}^{+\infty} \Pr\{X > j\}$. In the breaking of the infinite sum, note that both sum are divergent as $F_X(j) \to 1$ when $j \to \infty$. You may try $\Pr\{X = j\} = [1 - F_X(j-1)] - [1 - F_X(j)] $ in you step. This time the survival function tends to zero and the series break should be convergent. $\endgroup$ – BGM Sep 27 '17 at 2:16
  • 1
    $\begingroup$ As both sums are divergent, you are not able to do algebra with them as you did. $\endgroup$ – Joaquin San Sep 27 '17 at 2:27

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