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 $0.4<P(Y>1)<0.5$.

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.

  • $\begingroup$ 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. $\endgroup$ – Phil Mar 20 at 13:22
  • $\begingroup$ 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 $\sigma_z=0$. $\endgroup$ – C Monsour Mar 20 at 13:25

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