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Can You Find ... Random Variable Edition
Age 16 to 18
Challenge Level 
(a) ... the mean (expectation) is 2, the variance is 1, and $X$ takes only positive (integer) values? Can you find another? And a very different example?
- $\mathrm{P}(X=1)=\mathrm{P}(X=3)=\frac{1}{2}$, for example.
- If $Y$ has Poisson distribution with parameter 1, so $\mathrm{E}(Y)=\mathrm{Var}(Y)=1$, then $X=Y+1$ does what we want.
- If $Y$ has a geometric distribution with $p=\frac{1}{2}$ (so $Y$ can take values 1, 2, 3, ...), then $\mathrm{E}(Y)=\mathrm{Var}(Y)=2$. We can then modify this by increasing the probability of getting 2 itself to reduce the variance. So define $X$ by $$\mathrm{P}(X=x) = \begin{cases} \tfrac{1}{2}+\tfrac{1}{2}\mathrm{P}(Y=2) & \text{if $x=2$} \\ \tfrac{1}{2}\mathrm{P}(Y=x) & \text{otherwise} \end{cases}$$ and this can be calculated to work.
- $\mathrm{P}(X=0)=\mathrm{P}(X=4)=\frac{1}{4}$, $\mathrm{P}(X=2)=\tfrac{1}{2}$ for example; this can be generalised to $\mathrm{P}(X=2-k)=\mathrm{P}(X=2+k)=\frac{1}{k^2}$, $\mathrm{P}(X=2)=1-\tfrac{1}{k^2}$
- If $Y$ has a binomial distribution, $Y\sim \mathrm{B}(n,p)$, so that $Y$ takes the value 0, we can mix this with an increased probability for 2 as follows (this was found by playing around with possibilities): take $n\ge 4$, $p=\frac{2}{n}$ and $p'=\frac{1}{2(1-p)}=\frac{n}{2n-4}$. Then if we define $X$ by $$\mathrm{P}(X=x) = \begin{cases} (1-p')+p'\mathrm{P}(Y=2) & \text{if $x=2$} \\ p'\mathrm{P}(Y=x) & \text{otherwise,} \end{cases}$$ we find that $X$ has the required properties. In the case that $n=4$, we get $Y=X$, as this binomial distribution works.
- The mean of $X$ is $$\mathrm{E}(X)=\sum_{n=1}^\infty n.\mathrm{P}(X=n).$$ So we need a sum which is infinite. But we also require $\sum_{n=1}^\infty \mathrm{P}(X=n)=1$, so the sum can't be too "dramatically" infinite. If we, for example, took $\mathrm{P}(X=n)$ for each $n$, then the sum for $\mathrm{E}(X)$ would indeed sum to infinity, but the sum of the probabilities would also be infinite. If we took $n.\mathrm{P}(X=n)=1$ for each $n$, then the sum for $\mathrm{E}(X)$ would be infinite, but the sum of probabilities is $\sum_{n=1}^\infty \frac{1}{n}$. In Harmonically, we discovered that this sum is also infinite (though - in some sense - only just so: it grows towards infinity really really slowly). So this won't work either. But what about taking $n.\mathrm{P}(X=n)=\frac{1}{n}$? Then the sum for $\mathrm{E}(X)$ is infinite, but the sum of probabilities is $\sum_{n=1}^\infty\frac{1}{n^2}$. This is finite (see A Swiss Sum) so we are almost done: if this finite sum is $k$, then we can take $\mathrm{P}(X=n)=\frac{1}{kn}$ so that the probabilities sum to 1 but the mean is infinite.
- We can extend this idea by taking $\mathrm{P}(X=n)=\dfrac{1}{kn^s}$ for any $s$ with $1< s\le 2$, where $k$ is chosen to make the sum of probabilities equal 1; then the expectation will again be infinite. We don't even need to use the same value of $s$ for each $n$, though there are some restrictions on how quickly $s$ can approach 1 in such a case.
- The idea is the same as the previous part. If we choose $\mathrm{P}(X=n)=\dfrac{1}{kn^3}$, then the mean will be finite but the variance will be infinite. Now modifying it slightly by setting $\mathrm{P}(X=-1)=p$ and $\mathrm{P}(X=n)=\dfrac{1-p}{kn^3}$ for $n\ge1$, by choosing $p$ appropriately, we will end up with a mean of 0 but an infinite variance.
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