Let $X$ be a non-negative random variable, and let $Y \sim \chi^2_n / n$ (so that $E(Y) = 1$). $X$ and $Y$ are independent. Note that $X$ and $X\cdot Y$ have the same mean, while $X\cdot Y$ has larger variance. Is it true that for every $t > E(X)$, $$ \Pr(X > t) < \Pr(X\cdot Y > t)\,? $$

My special case is $X = z^TKz/z^Tz$, when $z\sim \mathcal{N}(0,I)$ and $K$ is a symmetric positive definite matrix, although I suspect this hold generally.

  • 1
    $\begingroup$ Sorry if I'm being dense, but how do you know that X and X*Y have the same mean? $\endgroup$ May 1, 2017 at 19:05
  • $\begingroup$ Due to their independence, $E(X\cdot Y) = E(X) \cdot E(Y) = E(X)$. $\endgroup$
    – R S
    May 2, 2017 at 14:45

3 Answers 3


A counter-example:

Let X be the distribution with 99% of its probability mass at 100, and the rest of its probability mass at 0. Let t be 99.5.

In cases where the realized value of X is 0, multiplication by Y will never result in a product above 99.5. (This is essentially true even if 1% of the probability mass concentrates slightly above zero, rather than exactly on zero). In cases where the realized value of X is 100, multiplication by Y will frequently result in a product less than 99.5.

  • $\begingroup$ In that case, what properties of X are needed for it to be true? And for the specific X in question? $\endgroup$
    – R S
    May 2, 2017 at 18:55
  • $\begingroup$ @RS The property doesn't seem to hold for you special case of X too. $\endgroup$
    – ab90hi
    May 5, 2017 at 12:55

This property doesn't hold true for all non-negative distributions of $X$.

Consider the case $X \sim \text{Bernouli}(p)$, for some $0<p<1 \implies E(X)=p$

and $Y \sim \chi^2(1)$

For $t\ \text{such that, }\ p<t<1$, $P(X>t) = P(X=1) =p$

$P(X.Y>t) = P(Y>t/X=1)*P(X=1) = P(Y>t)*p<p$

$\implies P(X>t) > P(X.Y>t)\\$


Update on the special case of X

$X = z'Kz/z'z$ where $z \sim N(0,I)$ and $K$ is positive definite

$\text{So K can be written as,} \\ K = U'DU \text{ where } U \text{ is orthogonal and } D \text{ is diagonal matrix with } d_i>0 \ \forall i$

$ \implies X=z'U'DUz/z'U'Uz =\sum_i d_iv_i^2/\sum_iv_i^2 \text{, where } V=Uz \sim N(0,I)$

$X = \sum_id_iv_i^2/\sum_i v_i^2,$ where $v_i^2 \sim \chi^2(1)$

Let us define $w_i = v_i^2/\sum v_i^2 \implies w_i \sim Beta(1/2,(n-1)/2)$

$E(w_i) = 1/n \implies E(X) = (\sum_{i=1}^n d_i)/n$

Since $w_i$'s are not independent it gets a bit complicated to derive the closed form distribution of X.

For simplicity let us look at the case when K is 2x2 matrix and D = Diagonal$(d_1,d_2)$

$X = d_1 w_1 + d_2 (1-w_1) = d_2 + (d_1 - d_2)*w_1 $, where $w_1 \sim Beta(1/2,1/2)$

$Y \sim Gamma(n/2,n/2)$

A Contradicting Example for special case of X

$Y \sim Gamma(1/2,1/2),\ \ Median(Y) \approx 0.47$

$X = d_1 + (d_2 - d_1)W,$ let $d_1=0.2,d_2=0.3 \implies X \in [0.2,0.3],$ $Median(X) = 0.25$

$XY \le 0.3Y \implies Median(XY) \le 0.3*Median(Y) \le 0.14$

Now since $E(X) = 0.25,$ consider $t=0.25 + \epsilon >E(X),$ for some small $\epsilon > 0.$ Also, $t > Median(XY)$

$P(X>t) \approx 0.5$

$P(XY>t) < P(XY>Median(XY)) = 0.5 \implies P(XY>t) < 0.5 $

$P(XY>t) < P(X>t),$ this example disproves it even for your special case as well.

I have ran a few simulations with different values of $K$ and found a few more contradicting cases.

  • $\begingroup$ The Bernoulli case is instructive, thanks. About my special case, I think that according to your notations, $Y\sim\Gamma(1,1)$ and not $\Gamma(1/2, 1/2)$. When I directly simulate your [0.3, 0.2] scenario, I still get that this property holds. $\endgroup$
    – R S
    May 8, 2017 at 16:56
  • $\begingroup$ $Y \sim \chi^2(n) / n $ for n=1 it does follow $Gamma(1/2,1/2)$ right? $\endgroup$
    – ab90hi
    May 8, 2017 at 17:58
  • $\begingroup$ In your example $n=2$, no? It's also not true that $Median(Y) \approx 0.47$ (this is true for 1 d.f of chi-square, but not two). But on further checking when I'm not as tried, it's true this example disproves it. Thanks! $\endgroup$
    – R S
    May 9, 2017 at 6:21

I see the intuition behind your question, but I'm not sure this holds for the general case.

First, you can re-write your original inequality as:

$$ \Pr(X < t) > \Pr(X\cdot Y < t) $$

This is equal to:

$$ F_X(t) > F_{X\cdot Y}(t) $$

Which is:

$$ \int_{0}^{t} f_X(i)di > \int_{0}^{t} f_X(i)g_Y(i)di $$

where $f_X(i)$ is the PDF of $X$, and $g_X(i)$ is the PDF of $Y$. You know the latter but not the former. As such, I see no way you can proof the above. Maybe you could think the proof would come from some type of monotonicity property of the Chi-Squared distribution, but $g_Y(i)$ is not monotonic for $n>2$.

In fact, you can very easily think of two distributions with same mean but different variance such that the accumulation of mass in the tails as be move up along the domain from the mean might be faster or slower, depending on the kurtosis (e.g. for normal).

Naturally, I haven't proved anything, nor given you a counterexample, so I might be wrong.

  • $\begingroup$ Thanks. A minor correction, I think the RHS should include $g_Y(t-i)$ instead of $g_Y(i)$. $\endgroup$
    – R S
    May 8, 2017 at 16:58
  • $\begingroup$ Mmm, why is that? Notice I reversed the original inequality, so my analysis is comparing the lower tails, which is the same problem, as both are assumed to have equal mean. $\endgroup$
    – luchonacho
    May 8, 2017 at 17:21

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