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For two nonnegative independent r.v.'s, $X,Y$, with the same distribution and finite second moment, I'm trying to show that $Var[\min(X,Y)]\leqslant Var[X]$.

Attempt 1. For the continuous case, I've written the first and second moments of $\min(X,Y)$ in terms of $X$ but have no idea how to proceed with them. Specifically, with $Z=\min(X,Y)$, I have

$\mathbb{E}\left[Z\right]=2\mathbb{E}[X]+\int_{0}^{\infty}F^2_{X}(z)dz$,

$\mathbb{E}\left[Z^2\right]=2\mathbb{E}\left[X^{2}\right]+2\int_{0}^{\infty}zF_{X}^{2}(z)dz$,

(where $F_{X}$ is the cdf of $X$) but don't know what to do with the integrals in the RHS's of the above expressions.

Attempt 2. Noting that $\min(X,Y)=\frac{1}{2}\left(X+Y-|X-Y|\right)$, I can write

\begin{equation} \label{eq1} \begin{split} Var\left[\min(X,Y)\right] & = Var[X]-\frac{1}{4}\left(\mathbb{E}\left[\left|X-Y\right|^2\right]+4\text{Cov}\left(X,|X-Y|\right)\right), \\ \end{split} \end{equation}

but am struggling to show that the RHS's second term (1/4(...)) is less than or equal to zero.

Any suggestions about how I might proceed or confirmation these are dead-ends would be appreciated.

ben
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  • @OliverDiaz Thanks for the response. Showing that $Var\left(X_1,\cdots,X_n\right)$ is nonincreasing for increasing $n$ indeed would be sufficient but I don't see how this is established in Lee David Chung Lin's answer to the linked post. If you have any insights into how it does establish this fact, please do share. – ben May 07 '21 at 03:12
  • Great question, and I don't have a solution yet, but it wouldn't surprise me if you have to use the following trick: if $\eta \in \mathbb{R}$ and $Z = \min(X, Y)$ then $$\mathbb{E}\left[\left(Z - \mathbb{E}[Z] \right)^2\right] \leq \mathbb{E}\left[(Z - \eta)^2 \right]\text{.}$$ – Clarinetist May 07 '21 at 13:33
  • @Clarinetist Thanks for responding. Would you mind explaining how the equality you've provided above is true (or pointing me to where I can find out)? – ben May 07 '21 at 13:53
  • @ben $$\begin{align} \mathbb{E}[(Z - \eta)^2] &= \mathbb{E}[(Z - \mathbb{E}[Z] + \mathbb{E}[Z] - \eta)^2] \ &= \mathbb{E}[(Z - \mathbb{E}[Z])^2] + \mathbb{E}[ (\mathbb{E}[Z] - \eta)^2] \end{align}$$ The only part of this equality depending on $\eta$ is $\mathbb{E}[ (\mathbb{E}[Z] - \eta)^2]$, which is nonnegative and thus minimized when the expectation is $0$, or $\eta = \mathbb{E}[Z]$. – Clarinetist May 07 '21 at 14:04
  • @Clarinetist Thank you. That makes sense. – ben May 07 '21 at 14:26

3 Answers3

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Note that $$ \text{var}[\min\{X,Y\}]{=\text{var}[\min\{X,Y\}|X<Y]\Pr\{X<Y\} \\+\text{var}[\min\{X,Y\}|X\ge Y]\Pr\{X\ge Y\} \\= \text{var}[X|X<Y]\Pr\{X<Y\}\\+\text{var}[Y|X\ge Y]\Pr\{X\ge Y\} . } $$Assuming $\Pr\{X=Y\}=0$, due to symmetry we have $\text{var}[X|X<Y]=\text{var}[Y|Y<X]$ and consequently $$ \text{var}[\min\{X,Y\}]=\text{var}[X|X<Y], $$ which leaves us with an obvious inequality $$ \text{var}[X|X<Y]\le\text{var}[X], $$ since reducing the domain of a random variable also reduces its variance (a formal proof comes from definition).

Mostafa Ayaz
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$Var(min(X,Y))=0.25 Var(X+Y-|X-Y|) = 0.25 (2Var(X)+Var(|X-Y|))-0.5 E((X+Y)|X-Y|)+0.5(E((X+Y)E(|X-Y|))$
$=0.5 Var(X)+0.25 Var(|X-Y|)-0.5 E((X+Y)|X-Y|)+0.5(E((X+Y)E(|X-Y|))$ Assume $E((X+Y)|X-Y|) \geq (E((X+Y)E(|X-Y|))$ which is true for uncorrelated RV as here $X+Y,X-Y$ are uncorrelated.
$\leq 0.5 Var(X)+0.25 E((X-Y)^2)-0.25 E(|X-Y|)^2$ $=0.5 Var(X)+0.25 (2 Var(X))-0.25 E(|X-Y|)^2 \leq Var(X)$

Lee David Chung Lin
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  • Thanks for the response. This resembles an attempt I made earlier today. But from what I can gather $X,Y$ independent does not imply $X+Y, X-Y$ independent (in general). https://math.stackexchange.com/a/782866/742588 – ben May 07 '21 at 14:35
  • Just because $X$ and $Y$ are independent, $|X+Y|$ and $|X-Y|$ need not be independent. As in, for example, $X,Y$ iid $U[0,1]$. – kimchi lover May 07 '21 at 14:36
  • Since we only worry about variance, uncorrelated would be enough ? which is true here. –  May 07 '21 at 14:38
  • I edited the proof. –  May 07 '21 at 14:41
  • @Balajisb Thank you! Just realised uncorrelatedness is sufficient. – ben May 07 '21 at 14:52
  • I modified the proof to make it clear. –  May 07 '21 at 14:54
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I have an intuitive way to see why this inequality holds.

In fact, you can calculate the PDF of $Z$ to be $f_Z(z)=2f_X(z)(1-F_X(z))$.

It means:

$$\begin{align} f_Z &>f_X & &\text{if} & z &< \mathrm{median}(X)~, \\ f_Z &< f_X & &\text{if} & z &> \mathrm{median}(X)~. \end{align}$$

so, the density of $Z$ is more compact compared to density of $X$. Hence, the variance of $Z$ is lower.

Lee David Chung Lin
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林胤均
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