# Setting bound for an infinite expected value

Say $X=2^Z$ and $Z$ is a geometric random variable with $p=1/2$.

It follows that, $E[X] = \infty$

So setting the upper bound by the markov inequality, $$P(X \geq t) \leq \frac{E[X]}{t} = \frac{\infty}{t}$$ for some value $t$.

However, this doesn't make much sense since we know the maximum probability is $1$.

Any advice on how to estimate the upper bound probability $P(X \geq t)$ ?

• You want to estimate the probability $\mathbb P(X\geqslant t)$ if the expectation of $X$ is infinite, isn't it? Do you have other information? – Davide Giraudo Jan 21 '15 at 14:22
• Yes I'd like to estimate the probability, $X = 2^Z$ where $Z$ is a geometric random variable with $p=1/2$. Hence why $E[X] = \infty$ – johnny Jan 21 '15 at 14:26
• @johnny: it is better to add it to the problem, because in such a formulation it does not make much sense. If we only know that the expected value of a r.v. $X$ is infinite, we cannot say anything on the rate of convergence to zero of $\int_{x}^{+\infty}f(t)\,dt$, where $f$ is the pdf of $X$. But in your case, you know the distribution of $X$. – Jack D'Aurizio Jan 21 '15 at 14:51
• @JackD'Aurizio updated, thanks :) – johnny Jan 21 '15 at 14:56

The Markov inequality still holds, it just isn't useful here. For your specific example though, $Z$ takes values in $1,2,3,\ldots$ so $X$ takes values in $2,4,8,\ldots$, and $$\mathbb P(X=k)=\mathbb P(2^Z = k)=\mathbb P(Z=\lg k)=\left(\frac12\right)^{\lg k}, k=2,4,8,\ldots.$$ (where $\lg$ denotes base-$2$ logarithm). If $t>0$, let $n=\min\{k\geqslant 1 : \lg k\geqslant t \}.$ Then $$\mathbb P(X\geqslant t)=\sum_{k=n}^\infty\mathbb P(X=2^k)=\sum_{k=n}^\infty \left(\frac12\right)^k=\left(\frac12\right)^{n-1}.$$

• Great answer, but shouldn't the probability depend on $t$ ? In other words, shouldn't $P(X>2)$ be different from $P(X>3)$ ? thanks – johnny Jan 21 '15 at 15:03
• Yes, $n$ is the smallest positive integer greater than or equal to $2^t$. – Math1000 Jan 21 '15 at 15:35