I'm trying to prove, or find a counterexample, for the following problem:

Let $Y = \{y_i\}_{i=1}^n$ be a set of data, where $y_i \ge 1$ for $i \in \{1,\ldots,n\}$, and let $\alpha$ be a natural number. Then the following holds between the variances of $Y$ and $Y^\alpha = \{y_i^\alpha\}_{i=1}^n$:

$Var[Y] \le Var[Y^\alpha]$

Note that:

$Var[Y] = \frac{1}{n}\sum_{i=1}^n(y_i - E[Y])^2$

$Var[Y^\alpha] = \frac{1}{n}\sum_{i=1}^n(y_i^\alpha - E[Y^\alpha])^2$

I tried a lot of examples, and a term-wise comparison always resulted in the following (for all $i \in {1,\ldots,n}$):

$(y_i - E[Y])^2 \le (y_i^\alpha - E[Y^\alpha])^2$

I'm still not sure if that's the case, but since I couldn't find any counterexamples, I used the identity $A^\alpha - B^\alpha = (A-B)(A^{\alpha-1} + A^{\alpha-2}B + \cdots + B^{\alpha-1})$ to expand left-hand side of the above inequality:

$(y_i^\alpha - E[Y^\alpha])^2 = (ny_i^\alpha - (y_1^\alpha+\cdots+y_n^\alpha))^2/n^2 = [\sum_{1\le j \le n}(y_i^\alpha - y_j ^\alpha)]^2/n^2 = [\sum_{1\le j \le n}(y_i-y_j)(y_i^{\alpha-1} + y_i^{\alpha-2}y_j + \cdots + y_j^{\alpha-1})]^2/n^2$

and the value $(y_i^{\alpha-1} + y_i^{\alpha-2}y_j + \cdots + y_j^{\alpha-1})$ is at least $\alpha$, since we assumed that $y_i$'s are at least 1.

But that's pretty much it, and I couldn't go any further, since the value $(y_i-y_j)$ might be negative.

Any help is appreciated. Please note that my approach might be totally wrong, so please feel free to use any suitable approach.

Edit: I finally found an example which shows the term-wise approach does not work. Let $Y=\{2,7,9.7\}$, and consider $\alpha = 2$. Then, $E[Y] = 6.23$ and $E[Y^2] = 49.03$, and for $y_2$ (which is 7) we have:

$(y_2 - E[Y])^2 = 0.59$, which is greater than $(y_2^2 - E[Y^2])^2 = 0.0009$.

Still, note that the inequality in question holds; i.e. $Var[Y] \le Var[Y^\alpha]$.


1 Answer 1


Somewhat more generally, consider a random variable $Y \ge 1$. $\text{Var}(Y^\alpha) = E[Y^{2\alpha}] - E[Y^\alpha]^2$. Take the derivative of this with respect to $\alpha$, noting that $\frac{d}{d\alpha} y^\alpha = (\ln y) y^\alpha$: $$ \frac{d}{d\alpha} \text{Var}(Y^\alpha) = E[2 \ln(Y) Y^{2\alpha}] - 2 E[Y^\alpha] E[\ln(Y) Y^\alpha] = 2 \text{Cov}(Y^\alpha, \ln(Y) Y^\alpha)$$

But since $y^\alpha$ and $\ln(y) y^\alpha$ are both nonnegative increasing functions of $y \ge 1$, $Y^\alpha$ and $\ln(Y) Y^\alpha$ are positively correlated, so this derivative is positive. Thus $\text{Var}(Y^\alpha) - \text{Var}(Y^1) \ge 0$.


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