# Probability depending on variance-to-mean ratio for a sum of normal distributions

let us consider three normally distributed random variables $$X \sim N(\mu_X,\sigma_X^2)$$, $$Y \sim N(\mu_Y,\sigma_Y^2)$$, $$Z \sim N(\mu_Z,\sigma_Z^2)$$, all of which being (mutually) stochastically independent. Moreover, all of them have strictly positive mean values, and $$\sigma_Y^2/\mu_Y \ge \sigma_Z^2/\mu_Z$$.

Let $$\alpha \in (0,0.5)$$ be arbitrary but fixed, and let $$\mathbb{P}[X>1] \le \alpha$$, $$\mathbb{P}[Y>1] \le \alpha$$, and $$\mathbb{P}[Z>1] \le \alpha$$.

My question is: If $$\mathbb{P}[X+Y>1]> \alpha$$, can we have $$\mathbb{P}[X+Z>1] \le \alpha$$?

(In a more abstract way: Can the considered probability decrease when $$Y$$ is replaced by a random variable $$Z$$ with lower or equal variance-to-mean ratio?)

Yes, we can have that situation.

Since the probabilities are continuous in the parameters, you can see this by taking a situation on the boundary and then by moving the parameters off the boundary by small amounts.

So we start by taking

$$\mu_x=\mu_z=0.25 \\ \mu_y=1.0 \\ \sigma_x=\sigma_z=0 \\ \sigma_y=1$$

Then $$P(X>1)=P(Z>1)=P(X+Z>1)=0$$, $$P(Y>1)=0.5$$, and $$P(X+Y>1)>0.5$$.

By continuity we can make sufficiently small upward adjustments to $$\sigma_x$$ and $$\sigma_z$$ and a sufficiently small downward adjustment to $$\mu_y$$, and have $$P(X>1)$$,$$P(Z>1)$$, and$$P(X+Z>1)$$ all less than $$0.1$$, $$P(X+Y>1)$$ still greater than $$0.5$$ and $$0.41)<0.5$$.

Also, you can make these changes sufficiently small that you still have $$\frac{\sigma_z^2}{\mu_z} < \frac{\sigma_y^2}{\mu_y}$$

Now take $$\alpha$$ between $$P(Y>1)$$ and $$0.5$$, and you are done.

• Thanks. I did not think about $\sigma_Z=0$ as a possibility. Then, we could even say that whenever $\mathbb{P}[X>1] < \alpha$ holds, we can choose $\mu_Z>0$ so small that it can be added to $X$ without violating the condition $\mathbb{P}[X+Z>1] \le \alpha$. This would also respect the inequality between the variance-to-mean ratios. – Phil Mar 20 at 13:22
• You don't actually need $\sigma_z=0$, and you notice I use continuity to adjust it to a small positive number, but it's a lot easier to think of the answer by starting with $\sigma_z=0$. – C Monsour Mar 20 at 13:25