# 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 $$\mu_Y$$.

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.

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