# What is the largest possible variance of a random variable on $[0; 1]$?

What is the largest possible variance of a random variable on $$[0; 1]$$?

It is evident that it does not exceed $$1$$, but I doubt, that $$1$$ is actually possible. The largest variance, for which I found the example is $$\frac{1}{4}$$. That is the variance of a random variable $$X$$, such that $$P(X = 1) = P(X = 0) = \frac{1}{2}$$, but I doubt that it is the largest possible one.

Why is it interesting:

Initially I wanted to find the largest possible second moment of $$X - Y$$, where $$X$$ and $$Y$$ are i.i.d. random variables on $$[0; 1]$$. Then I found: $$E(X - Y)^2 = E(X^2 - 2XY + Y^2) = EX^2 - 2EXY + EY^2 = 2(EX^2 - EXY + EXY - EXEY + Cov(X, Y)) = 2(EX^2 - {(EX)}^2) = 2VarX$$ And thats where I am now.

This question is partially inspired by: Probability distribution to maximize the expected distance between two points

• BTW, the "motivation" part can be derived more easily: $E[X-Y]^2 = Var(X-Y) + E[X-Y]^2$, but since $X, Y$ are i.i.d., $E[X-Y]=0$ and $Var(X-Y) = 2Var(X)$. – antkam Feb 9 at 21:11

The answer is $$1/4$$. For any probability distribution on $$[0,1]$$ the point $$p=(EX,EX^2)$$ will be a point in the convex hull of the set $$S = \{(x,x^2):x\in[0,1]\}$$. (This is a segment of a parabola.) The variance is the height of $$p$$ above the set $$S$$. This is clearly maximized when $$p$$ lies on the straight line connecting $$(0,0)$$ to $$(1,1)$$. By calculus, this is attained at $$p=(1/2,1/2)$$, which is $$1/4$$ above the point $$(1/2,1/4)\in S$$.

That $$p$$ is in the convex hull of $$S$$ is a consequence of Caratheodory's theorem: each element $$p$$ of the convex hull of $$S$$ is a weighted combination of at most 3 elements of $$S$$: Let that combination be $$p=\sum_{i=1}^3 w_i\cdot(x_i,x_i^2)$$, where the $$w_i\ge0$$ add up to $$1$$. Now look at the prob distribution for which $$PX=x_i)=w_i$$. Its first 2 moments are the components of $$p$$.

In the special case at hand Caratheodory's theorem is trivial. Every point in the convex hull of $$S$$ is on a chord of $$S$$. If $$p$$ is already in $$S$$, it is of the form $$p=(x,x^2)$$, and the prob law $$P(X=x)=1$$ does the trick. Otherwise, the line passing through $$(0,0)$$ and $$p$$ cuts $$S$$ at $$q$$; the chord in question can be between $$(0,0)$$ and $$q$$, and $$a$$ can be chosen so $$P(X=(0,0))=a, P(X=q)=1-a$$ does the trick.

More generally, as a comment suggests, this is a fundamental property of a expectation operator. An expectation $$ET(X)$$ of a vector valued function is a particular weighted average of the possible values of the function $$T(x)$$. A probability law, one can think, amounts to a choice of weights. In problems like this one, the set of possible values of $$ET(X)$$ you get as you vary the probability law of $$X$$ is the convex hull of the set of values of the function $$T(x)$$.

• perhaps i'm missing something obvious, but why is the point $p$ in the convex hull of $S$? presumably this is a consequence of the Expectation operator, but i cant see an obvious proof. can you help? – antkam Feb 9 at 19:58
• @antkam I have edited my answer. – kimchi lover Feb 9 at 20:28
• thanks! that theorem is exactly what i was looking for. very neat, especially the geometric interpretation. however, i must say, the theorem is much stronger than the requested result. i have written up a more elementary (but less insightful) proof as alternative answer. – antkam Feb 9 at 20:39

Another elementary proof: $$X^2\leq X$$ since $$X\in[0,1]$$. Therefore $$Var(X)=\mathbb E[X^2] - \mathbb E[X]^2\leq \mathbb E[X]-\mathbb E[X]^2 = z-z^2 \leq \frac14.$$ The last indequality valid since function $$f(z)=z-z^2$$ reaches its maximal value $$\frac14$$ at point $$z=\frac12$$.

To reach equality in the first inequality $$\mathbb E[X^2]\leq \mathbb E[X]$$ we need $$X^2=X$$ a.s. This is possible only when $$X$$ takes values $$0$$ and $$1$$. The second indequality $$z-z^2\leq \frac14$$ is equality if $$z=\mathbb E[X]=\frac12$$.

So the upperbound is $$\frac14$$ and it is achieved by $$\mathbb P(X=0)=\mathbb P(X=1)=\frac12$$ only.

A more elementary proof:

Variance is shift-invariant, i.e. let $$Y = X - 1/2$$ and $$Var(Y) = Var(X)$$. So we just need to consider $$Y$$ on $$[-1/2, +1/2]$$. The requested result follows from these observations:

• $$Y^2 \in [0, 1/4] \implies E[Y^2] \in [0, 1/4] \implies E[Y^2] \le 1/4$$

• $$E[Y]^2 \ge 0$$ since it is a square.

• (In fact, $$Y \in [-1/2, 1/2] \implies E[Y] \in [-1/2, 1/2] \implies E[Y]^2 \in [0, 1/4]$$.)
• $$Var(Y) = E[Y^2] - E[Y]^2 \le 1/4 - 0 = 1/4$$

This proves the requested upperbound of $$1/4$$; which the OP has already shown is achievable. This also shows that achieving this bound requires $$E[Y] = 0$$ and $$E[Y^2]=1/4$$ which in turn requires $$Y^2 = 1/4$$ which in turn requires $$Y = \pm 1/2$$. The OP's example is obviously the only such distribution.