$\mathbb{E}$[$X_{\mathbb{E}(T)}]$

Question

Consider an experiment consisting of a sequence of trials, each of which yields a success or a failure. Let $X_{n}$ be the number of successes after n trials, and let $T$ be the expected number of trials for the first success to occur.

Then I propose the following conjecture:

$\mathbb{E}$[$X_{\mathbb{E}(T)}$] $\leq$ $1$

We have equality when the probability of success is fixed (does not change from trial to trial). If we let this probability be $p$, then $\mathbb{E}(T) = \frac{1}{p}$, and so $\mathbb{E}$[$X_{\mathbb{E}(T)}] = \mathbb{E}$[$X_{\frac{1}{p}}] = p.\frac{1}{p} = 1$

We can also achieve $\mathbb{E}$[$X_{\mathbb{E}(T)}$] < $1$, as in the following experiment: Let a trial consist of throwing a fair six-sided die, and let a success be obtaining an outcome that we have obtained before (a duplicate). Then:

$\mathbb{E}(T) = 1.0 + 2.\frac{1}{6} + 3.\frac{5}{6}.\frac{2}{6} + 4.\frac{5}{6}.\frac{4}{6}.\frac{3}{6} + 5.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{4}{6} + 6.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{5}{6} + 7.\frac{5}{6}.\frac{4}{6}.\frac{3}{6}.\frac{2}{6}.\frac{1}{6}.1 \approx 3.77 $

However $\mathbb{E}[X_{3.77}] \leq \mathbb{E}[X_{4}] = 1.{4 \choose 2}.\frac{6.5.4}{6^4} + 2[{4 \choose 3}.\frac{6.5}{6^4} + \frac{1}{2}.{4 \choose 2}.\frac{6.5}{6^4}] + 3.\frac{6}{6^4} \approx 0.89 < 1$

Can anyone provide some insight into what is going on here? Is the conjecture true?

Thanks

Answer 1

Without more assumptions on the trials, this isn't true. Consider the following example:

Let $E_k$ be the event that trial #$k$ is a success. Let the joint distribution be such that $E_{1}, E_{4}, E_{9}, E_{16}, \dots$ are all independent with probability $1/2$, and $E_2, E_3$ are successes if and only if $E_1$ is a success, $E_5, E_6, E_7, E_8$ are successes if and only if $E_4$ is a success, etc. Then we note that the time $T$ of first success takes the value $n^2$ with probability $2^{-n}$, so $$ E[T] = \sum_{n \geq 1} n^2 2^{-n} = 6. $$ But we can also compute $E[X_6] = 0\cdot \frac14 + 3 \cdot \frac14 + 3 \cdot \frac14 + 6 \cdot \frac14 > 1$.