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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].

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    Thanks Louis. I guess I can only try and show it for my specific case then.2012-01-03

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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 $1-2m$.

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    @gallamine: I hope my addition is enough.2012-01-04
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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 $

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    Sasha, yes, you are right. But, I don't think I worded my question correctly. I will try and clarify it.2012-01-03