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Question

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 $(\mu_Y,\sigma_Y^2)$?


Thoughts

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 $\mu_Y$.

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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

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

Consequently,

$f_z$ 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.

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  • $\begingroup$ Wonderfully clear answer, thank you! $\endgroup$ – robust Sep 8 '19 at 7:29

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