Take two independent, non-degenerate Normals, $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$.

Define $Z=aX + Y$ with $a>0$.

This implies $Z \sim N(a\mu_X+\mu_Y,a^2\sigma_X^2+\sigma_Y^2)$

  1. If $Pr(Z\leq 0)$ decreases with $a$, what must be true about $\mu_X$ and $\mu_Y$?

  2. If $Pr(Z\leq z)$ with $z>0$ decreases with $a$, what must be true about $(\mu_X, \sigma_X^2)$ and $(\mu_Y,\sigma_Y^2)$?


Increasing $a$ has two effects on $aX$.

  1. It scales up the mean of $aX$.
  2. It scales up the variance of $aX$.

If $\mu_X>0$, then the first effect increase the mean value of $aX$, suggesting a decrease in $\Pr(aX\leq0)$.

However, with $\mu_X>0$ the second effect makes $\Pr(aX\leq0)$ increase with $a$. The intuition is that for a Normal an increase in the variance means more mass is away from the mean, which means more mass falls in the region below $\mu_X$ and hence also below $0$.

We also know that $\Pr(aX\leq0) = \Pr(X\leq0)$, and hence $aX$ is not affected by $a$. This means that effects 1 and 2 must be exactly offsetting each other.

Coming back to $Z$, we also know that $Y$ is not affected by $a$.

So we still have the same two effects of $a$ now working on $Z$ instead of $aX$. Intuitively, it seems to me that the first effect is as strong on $Z$ as on $aX$, while the second effect is attenuated by having the additional variance $\sigma_Y^2$. This suggests that if $Pr(Z\leq 0)$ decreases with $a$, then $\mu_X>0$ and we learn nothing about $\mu_Y$.


1 Answer 1


By choosing suitable units of measurement and working in terms of $a\sigma_X$ instead of $a,$ you can reduce the problem to the case $\mu_X=0,$ $\sigma_X=\sigma_Y=1.$

In that circumstance $Z-\mu_Y=aX+Y-\mu_Y$ has a Normal$(0,\sqrt{1+a^2})$ distribution, which stays centered at $0$ but whose spread increases with $a,$ making it clear that the chance $Z$ does not exceed $z$ increases when $z-\mu_Y$ is negative and--by symmetry--must decrease if and only if $z-\mu_Y$ is positive.

This reasoning translates into the following formal demonstration.

Letting $\Phi$ be the standard Normal CDF, define the function $f_z$ for positive arguments $a$ as

$$f_z(a)=\Pr(Z\le z) = \Phi\left(\frac{z - (a\mu_{X} + \mu_Y)}{\sqrt{a^2\sigma_X^2 + \sigma_Y^2}}\right).$$

Since the derivative of $\Phi$ is the standard Normal density, it is everywhere positive, whence (by the Chain Rule) the sign of $f^\prime(a)$ is the sign of the derivative of the argument of $\Phi,$ given by

$$\eqalign{ \frac{d}{da}\frac{z - (a\mu_X+\mu_Y)}{\sqrt{a^2\sigma_X^2 + \sigma_Y^2}}&=-\mu_{X}(a^2\sigma_X^2 + \sigma_Y^2)^{-1/2} \\&- \frac{1}{2}(a^2\sigma_X^2 + \sigma_Y^2)^{-1/2-1}(2a\sigma^2_X)(z - (a\mu_X+\mu_Y)) \\ &=\frac{-\mu_X(a^2\sigma_X^2 + \sigma_Y^2)-a\sigma_X^2(z-(a\mu_X+\mu_Y))}{(a^2\sigma_X^2 + \sigma_Y^2)^{3/2}}. }$$

Because the denominator is positive, the sign of the derivative is the sign of the numerator, which simplifies to

$$-\mu_X\sigma_Y^2 - a \sigma_X^2 (z-\mu_Y).$$


$f_z$ is decreasing at $a$ if and only if $$z \gt \mu_Y -\frac{1}{a} \frac{\mu_X\sigma_Y^2}{\sigma_X^2}.$$

When $f_z$ decreases for all $a,$ necessarily $z \ge \mu_Y,$ and conversely.

  • $\begingroup$ Wonderfully clear answer, thank you! $\endgroup$
    – robust
    Commented Sep 8, 2019 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.