# minimum cdf given mean and variance

Suppose I have some distribution with mean $\mu$ and variance $\sigma^2$. How to prove that for the distribution to have min $cdf$ over all the distributions with mean $\mu$ and variance $\sigma^2$, the distribution should be normal? I want to show that the cdf for normal distribution with mean $\mu$ and variance $\sigma^2$ (i.e. $\Phi_{\mu,\sigma^2}(x)$) is lower bound for all the cdf's of distributions with mean $\mu$ and variance $\sigma^2$.

More formally, let $F(x)$ be a cdf for some distribution with mean $\mu$ and variance $\sigma^2$. I want proof that $\Phi_{\mu,\sigma^2}(x) \leq F(x) \ \forall x$

• I don't quite follow this. What do you mean by "min $cdf$"? The minimum for a population that is truly distributed as a normal is $-\infty$, so that would be lower than any finite minimum by definition, but I don't know if that is what you are asking about. – gung - Reinstate Monica Jan 2 '16 at 21:50
• I edited the question with more formal details. – omdsoimoin Jan 2 '16 at 22:06
• Asking for such a lower bound makes so little sense that I wonder whether you might be confusing it with the differential entropy. – whuber Jan 2 '16 at 22:34
• Like @whuber I have no earthly idea what your edit might mean. We usually denote the CDF $F(x)$ (& for the normal distribution, $\Phi(x)$). By definition, the CDF outputs a value in $[0,1]$, whether the distribution is normal or not. The PDF is denoted $f(x)$ (/ $\phi(x)$ for a normal). It is most typically thought of as the derivative of the CDF. – gung - Reinstate Monica Jan 2 '16 at 23:38
• I think the question is perfectly clear and answerable as it stands. I don't see the difficulty here; the proposed result is obviously false -- an answer would consist simply of any of the (completely obvious) counterexamples. (Yes, it may be that the actual question is different from what was asked - so let's just answer this one and let the OP ask a new one if that turns out to be the case). I have an answer ready to go. – Glen_b Jan 3 '16 at 0:24

That this is not the case can easily be seen simply by computing the cdf for a different distribution than the normal.

So for example, consider a standard normal (i.e. let's fix $\mu$ and $\sigma$ at 0 and 1, without loss of generality$^\dagger$).

Let's try a uniform on $(-\sqrt{3},\sqrt{3})$ (which also has mean 0 and sd 1) for comparison:

We can see that it's not the case that $\Phi\leq F$ everywhere. There are places where either will exceed the other.

$\dagger$ we can do the same for the general $\mu,\sigma$ case by linearly scaling this example.

Indeed, if you're going to have the means be equal and you're going to make $F\gt\Phi$ in some region, it's going to have to be smaller somewhere else.

• Showing the contrapositive might be more convincing and more general: when $F$ and $G$ are (cumulative) distribution functions, $F$ has a finite expectation $\mu$, and $G(x)\le F(x)$ for all $x$, then $\mu$ (obviously) is a lower bound for the expectation of $G$ because the difference of expectations is the area enclosed between the graphs of those two functions. This immediately implies the O.P. is trying to prove the impossible. – whuber Jan 3 '16 at 23:05
• @whuber it would certainly be more general (and I'd say, more useful); but I don't see how it would be more convincing of the falsity of the claim per se since a single counterexample establishes that the claim is false. While writing it, I did think about modifying my final sentence along those lines but worried it would not be of sufficient benefit to the OP. I will think about adding that argument (but I plan to retain the simple counterexample in any case, since "just try an example" is an obvious thing for people to do, even if they can't come up with a general argument). – Glen_b Jan 4 '16 at 0:44