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Let the random variable X be distributed (not necessarily normally) with some mean $\mu$ and some variance $\sigma^2$. Under what conditions is it true that

$$ var(X|X<a) \leq var(X)$$ for some $a$ in the support of $X$? I can show it to be true for a normal random variable, and I believe it to be true for the Gumbel and logistic distributions (I just compute the variances in R for some different parameters). It is not true for the gamma distribution.

We can make assumptions on the support of $X$, or on, say, the log-concavity. But how do I guarantee the statement?

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2 Answers 2

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Without loss of generality, and simplicity of calculations, assume $E[X]=\mu=0$. Subtracting a constant from a random variable does not change its variance. Define Y to be the truncated random variable: $$f_y(y)= b f_x(y)$$ for $y<a$ and zero otherwise, where $b^{-1}=\int_{-\infty}^a f_x(x)dx\le1$. Then, $$E[Y^2] = \int_{-\infty}^a y^2 b f_x(y)\, dy = b\left(E[X^2]-\int_a^{\infty} y^2 f_x(y)\, dy\right)$$ $$E[Y] = \int_{-\infty}^a y b f_x(y)\, dy = b\left(E[X]-\int_a^{\infty} y f_x(y)\, dy\right)= -b\int_a^{\infty} y f_x(y)\, dy$$ so, $$V[Y]=E[Y^2]-E[Y]^2=bV[X] - b\int_a^{\infty} y^2 f_x(y)\, dy -b^2\left(\int_a^{\infty} y f_x(y)\, dy\right)^2$$ Define Z to be the random variable: $$f_z(z)= c f_x(z)$$ for $z>a$ and zero otherwise, where $c^{-1}=\int_a^{\infty} f_x(x)dx = 1-b^{-1}$. $$V[Y]=bV[X] - {b\over c}E[Z^2] -\left({b\over c}E[Z]\right)^2=bV[X] - (b-1)E[Z^2] -\left((b-1)E[Z]\right)^2$$ $$V[Y]=bV[X] - (b-1)V[Z] -b(b-1)E[Z]^2$$ $$\Delta=V[Y]-V[X]=(b-1)\left(V[X]-V[Z]-bE[Z]^2\right)$$ If $b=1$, then $V[Y]=V[X]$, otherwise $b>1$ in which case, $V[Y] \le V[X]$ iff: $$V[Z]+bE[Z]^2 \ge V[X]$$ where $Z$ is the random variable with a distribution corresponding to the removed part of $X$ for the case $\mu=0$. For the general case, the condition is: $$V[Z]+bE[Z-\mu]^2 \ge V[X]$$

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Here's a much quicker answer than the accepted one:

As in @Dean's solution, without loss of generality assume $E[X]= 0$.

Let

  • $X^+ = \begin{cases}X : X > a \\ 0 : \text{otherwise} \end{cases}$
  • $X^-=\begin{cases}X : X \le a \\ 0 : \text{otherwise} \end{cases}$

Clearly, $X = X^+ + X^-$. Also note that $X^+X^-=0$ (with probability $1$) because they cannot both be nonzero simultaneously.

\begin{align*} Var[X] &= E[X^2] - \underbrace{E[X]^2}_{=0} \\ &= E[(X^+ + X^-)^2]\\ &= E[(X^+)^2) + (X^-)^2 + 2 X^+X^-]\\ &= E[(X^+)^2)] + E[(X^-)^2] + 2 E[\underbrace{X^+X^-}_{=0}]\\ &= E[(X^+)^2)] + E[(X^-)^2]. \end{align*}

Now, note \begin{align*} Var[X^+] &= E[(X^+)^2] - E[X^+]^2\\ &\le E[(X^+)^2] \qquad \qquad \hspace{2em} \qquad \qquad \text{(since $E[X^+]^2\ge 0$)}\\ &\le E[(X^+)^2]+E[(X^-)^2]\qquad \qquad \qquad \text{(since $E[(X^-)^2]\ge 0$)}\\ &= Var[X]. \end{align*}

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  • $\begingroup$ Please see the comment directed to this answer below (in case it is deleted, it's here) $\endgroup$ Oct 9, 2021 at 14:12
  • $\begingroup$ @Arctic Char: Thanks for your comment. But just so I understand, do you agree with Dean's result ? I see no errors in Dean's argument ( Dean: thanks for a great step by step argument ) but I have no intuition on how a truncated variable's variance is not always less than the variance of the not truncated variable. Intuitively, it seems like if you narrow the support, then the new variance should always be less than the original variance. But I must be missing something when thinking about this. Thanks for any insights. $\endgroup$
    – mark leeds
    Oct 19, 2021 at 5:43
  • $\begingroup$ @Arctic Char you're right, my answer does not concern the truncated variable, but a function of the variable. It is indeed something different. My bad! $\endgroup$ Oct 20, 2021 at 18:09
  • $\begingroup$ @Zach Siegel: Do you know if this means that the other answer is correct. Thanks. $\endgroup$
    – mark leeds
    Oct 23, 2021 at 14:34

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