# Distribution function given expectation and maximal variance

I'm working on the following problem I got in a hw but I'm stuck. It just asks to find the distribution function of a random variable $X$ on a discrete probability spaces that takes values in $[A,B]$ and for which $Var(X) = \left(\frac{B-A}{2}\right)^{2}.$

I got that this equality gives the expected values $E(X) = \frac{A+B}{2}$ and $E(X^{2}) = \frac{A^{2}+B^{2}}{2}$, but I can't see why this gives a unique distribution (as the statement of the problem suggests).

I also found the distribution function $p(x) = 0$ for $x \in (A,B)$ and $p(A)=\frac{1}{2}$, $p(B)=\frac{1}{2}$ that works for example, but I don't see how this is the only one. Can anyone shed some light please?

Thanks a lot!

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If the question asks to find the distribution function of A random variable that blablabla, it's probably because there is not only one. To be quite honest I would be surprised if there was only one. – Patrick Da Silva Oct 7 '12 at 5:32
yes, I figured, but how could you characterize all of them knowing only the expectation and the variance? (I suppose this is what the question asks) – Dquik Oct 7 '12 at 5:46
maybe we can do it in this nice case when the variance has maximal value (we know that $Var(X) \leq \left(\frac{B-A}{2} \right)^{2}$ in general... – Dquik Oct 7 '12 at 5:47
I don't think you have any idea how ugly random variables can be. Knowing expectation and variance is very little. Now that I think about it maybe there's something to do... but I'm not very optimistic. – Patrick Da Silva Oct 7 '12 at 5:52
@PatrickDaSilva In principle you are right that random variables may be ugly, as you call them, but here, there is indeed a unique maximizing distribution, which is the discrete one you found in your answer. – Did Oct 7 '12 at 7:25

Let $m=\frac12(A+B)$ and $h=\frac12(B-A)$. The OP indicates in a comment how to prove that any random variable $X$ with values in $[A,B]$ and such that $\mathrm{Var}(X)=h^2$ is such that $\mathbb E(X)=m$ and $\mathbb E((X-m)^2)=h^2$.

Starting from this point, note that $|X(\omega)-m|\leqslant h$ for every $\omega$ since $A\leqslant X(\omega)\leqslant B$, hence $(X-m)^2\leqslant h^2$ everywhere. Together with the equality $\mathbb E((X-m)^2)=h^2$, this proves that $(X-m)^2=h^2$ almost surely, that is, $X\in\{A,B\}$ almost surely. Now, use once again the fact that $\mathbb E(X)=m$ to deduce that $\mathbb P(X=A)=\mathbb P(X=B)=\frac12$.

Edit: Let us recall why $Y\geqslant0$ almost everywhere and $\mathbb E(Y)=0$ imply that $Y=0$ almost everywhere.

Fix $\varepsilon\gt0$, then $Y\geqslant0$ almost everywhere hence $Y\geqslant\varepsilon\mathbf 1_{Y\geqslant\varepsilon}$ almost everywhere. This implies that $0=\mathbb E(Y)\geqslant\varepsilon\,\mathbb P(Y\geqslant\varepsilon)$, that is, $\mathbb P(Y\geqslant\varepsilon)=0$. Now, $[Y\ne0]=[Y\gt0]$ is the countable union over every positive integer $n$ of the events $[Y\geqslant1/n]$ hence $\mathbb P(Y\ne0)=0$. QED.

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Am I wrong saying this is a "non-scaled" version of my proof? It feels like it. – Patrick Da Silva Oct 7 '12 at 8:08
@PatrickDaSilva The idea is similar but the realizations of this idea seem different, even once one omits this scaling/unscaling thing, which is minor. Note for example that the method in my post nowhere assumes that the distribution has a density or is discrete or anything similar. Note also that you assume that the distribution has a density, only to reach the conclusion that it has not... :-) However, as I said, yes, in the end the idea is similar. – Did Oct 7 '12 at 8:14
Good, we agree =) I felt a little sketchy I admit. – Patrick Da Silva Oct 7 '12 at 8:16
If you ask me, I see no problem with sketchy solutions (as long as one recognizes them as such), nor with so-called stupid questions (which in my book simply do not exist...). – Did Oct 7 '12 at 8:22
I meant "trivial question" then. ;) – Patrick Da Silva Oct 7 '12 at 16:40

If we simplify the problem and scale things appropriately, you're essentially looking for a distribution function $f(x)$ such that $$\int_{-1}^{1} f(x) \, dx = 1, \quad \int_{-1}^1 x f(x) \, dx = 0, \quad \int_{-1}^1 x^2 f(x) \, dx = 1.$$ Let us look only at the first two integrals first, so that we only look at variables $X$ whose $f$ satisfy $$\int_{-1}^1 f(x) \, dx = 1, \quad \int_{-1}^1 x^2 f(x) \, dx = 1.$$ This implies that $$\int_{-1}^1 (1 - x^2) f(x) \, dx = 0.$$ Notice that $1-x^2$ is positive and $f$ is a positive measure on $[-1,1]$ for which $\int_{-1}^1 f = 1$. Consider $\varepsilon > 0$ and $A_{\varepsilon} = [-1+\varepsilon, 1 - \varepsilon]$. Assume that $\int_{A_{\varepsilon}} f(x) \, dx > 0$. Therefore $$0 = \int_{-1}^1 (1-x^2)f(x) \, dx \ge \int_{A_{\varepsilon}} (1-x^2) f(x) \, dx \ge \int_{A_\varepsilon} (1-(1-\varepsilon)^2) f(x) \, dx > 0$$ because the last integral is a constant times the integral over $A_{\varepsilon}$ of $f$. This contradiction shows that $\int_{A_{\varepsilon}} f(x) \, dx = 0$ for every $\varepsilon > 0$.

Since probability measures are continuous, we can let $\varepsilon \to 0$ and know that $p(]-1,1[) = 0$. This means $p( \{-1,1\} ) = 1$ by taking complements. Since the expectation has to be $0$, the variable has to be symmetric, i.e. $p(-1) = p(1) = \frac 12$.

Hope that helps,

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This answer is probably better off being a comment, but at least it makes me a little bit more optimistic that your random variable might be unique. I could "argue that all the weights have to be on the endpoints" using measure theory a little better but I didn't know if you knew measure theory. It wouldn't be that hard. – Patrick Da Silva Oct 7 '12 at 6:15
I forgot to say that the probability space is discrete, sorry – Dquik Oct 7 '12 at 6:17
anyway, please elaborate more, I'm not sure I understood the argument anyway (you can use measure theory) – Dquik Oct 7 '12 at 6:17
@Dquik : Probability space is discrete? That changes a lot! – Patrick Da Silva Oct 7 '12 at 6:34
@Dquik : I actually didn't need to assume the space was discrete! There literally is only one variable with such properties. I just needed to go through my assumptions a little bit more in detail. The whole proof is up there. Feel free to ask if you have any questions. – Patrick Da Silva Oct 7 '12 at 6:44