# prove variance of a RV of 1's and 0's is greater than var of a RV of values on (0,1]

I'm trying to prove that the variance of a RV whose values are discrete 1's or 0's is greater than the variance of a RV who's values are 0's or continuous on the domain (0,1], where any "1" in the Bernoulli RV corresponds to a value on (0,1] in the other RV. Intuitively, I think this is the case, but I'm trying to demonstrate it.

For instance, I know that $Var[X] > Var[\alpha X]$ where $0 \le \alpha < 1$, but what if $\alpha$ is a RV on the interval $(0,1]$?

The reason I'm asking is that I'm sampling a RV with values on the interval (0,1] and I'm using a binomial confidence interval as an upper bound (you can read more of the context from an old discussion here), and I need to prove that that's a reasonable upper bound on the confidence interval, since the values could all be 1's or 0's, but have the possibility of being between those values.

Edit Sasha provided a counterexample to my question as originally stated. The distribution in question is such that it's Bernoulli-like, however instead of having only 1's and 0's, the "true" values can take on values on the interval (0,1]. So, for Sasha's case of $p_{true} = 0.01$, the distribution in question would have a delta function at 0 of value $1-p_{true}$ and the rest of the distribution would be on (0,1].

• It's not true in general, since the always 1 random variable has lower variance than the uniform RV on [0,1]. – Louis Jan 3 '12 at 22:08
• Thanks Louis. I guess I can only try and show it for my specific case then. – gallamine Jan 3 '12 at 22:17

## 2 Answers

It is true if you add a condition that the means are the same. So if $X$ takes values in $[0,1]$ then $Var(X) \le E[X] - E[X]^2$ and if it is helpful then the latter is less than or equal to $\frac{1}{4}$.

Added: As a proof of this statement: since $0 \le X \le 1$ we have $X^2 \le X$ and so $E[X^2] \le E[X]$ and so $Var(X) = E[X^2] - E[X]^2 \le E[X] - E[X]^2$.

In this case with $X^2 \le X$, then $E[X^2] = E[X]$ if and only if $X^2 = X$ almost always, i.e. when $X=0 \text{ or } 1$ almost always, i.e. with a Bernoulli distribution.

$m - m^2$ is an increasing function of $m$ in the interval $[0, 1/2]$ since its derivative is $1-2m$.

• Henry, does this hold if the mean of the RV in question is less than or equal to the mean of the Bernoulli RV? Also, I'm a little confused by your notation - where does the 1/4 come from? – gallamine Jan 3 '12 at 23:48
• @gallamine: for the first part of your comment, yes if both means are less than $1/2$. – Henry Jan 3 '12 at 23:50
• @gallamine: for the second part of your comment, $1/4$ is maximum possible variance of a random variable taking values in $[0,1]$ and occurs for a Bernoulli RV with mean $1/2$. Ignore it if it is too big for your purposes. – Henry Jan 3 '12 at 23:54
• I'm trying to work out the proof for this, but I'm having trouble. Can you help me out a bit? – gallamine Jan 4 '12 at 0:11
• @gallamine: I hope my addition is enough. – Henry Jan 4 '12 at 0:25

As a counterexample, consider Bernoulli RV, $X \sim \mathrm{Be}(p)$ with $p=0.01$ and a uniform RV, $Y \sim \mathcal{U}(0,1)$: $$\mathbb{Var}(X) = p(1-p) = 0.0099 < \mathbb{Var}(Y) = \frac{1}{12} \approx 0.0833$$

• Sasha, yes, you are right. But, I don't think I worded my question correctly. I will try and clarify it. – gallamine Jan 3 '12 at 22:10