# Questions related to normal PDF and CDF

A random variable $$Y$$ has the Gaussian distribution on $$\mathbb{R}$$ with mean $$\mu$$ and variance $$\sigma^{2}$$. Select all the necessarily true statements from the following.

a. $$\mathbf{P}(Y \geq y) \leq \frac{1}{\sqrt{2 \pi \sigma}} \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \in \mathbf{R}$$

b. $$\mathbf{P}(Y \geq y) \leq \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \in \mathbf{R}$$

c. $$\mathbf{P}(Y \geq y) \leq \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \geq \mu$$ and $$\mathbf{P}(Y \leq y) \leq \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \leq \mu$$

d. $$\mathbf{P}(Y \geq y) \leq \frac{1}{\sqrt{2 \pi \sigma^{2}}} \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \geq \mu$$

e. $$\mathbf{P}(Y \geq y) \leq \exp \left(-(y-\mu)^{2} /\left(2 \sigma^{2}\right)\right)$$ for all $$y \geq \mu$$

Clearly a and b are false when $$y$$ is really close to $$-\infty$$, but for c, d, and e, my simulation shows they are correct but I need some help with theoretical solutions.

• Hint: consider how the right hand sides change as $\sigma^2$ varies from near $0$ to very large numbers; and note that if any of these statements are true, then they must be true for all $\sigma^2\gt 0.$
– whuber
Jan 21, 2021 at 19:50
• I see how the RHS vary for different $\sigma^2$, they are close to 0 is $\sigma^2$ is near 0 and close to 1 is $\sigma^2$ is very large. However, I could not relate this in comparison to the LHS. Jan 22, 2021 at 1:55
• When somebody claims that a number (such as any of the left hand sides of your inequalities) is always less than some variable value (any of the right hand sides), then it must be less than all possible values of the right hand side. Since you can choose $\sigma^2$ to make the right hand sides arbitrarily close to $0,$ the left hand sides cannot be positive. Draw the obvious conclusions.
– whuber
Jan 22, 2021 at 14:14

c is true. It suffices to prove it is true for the first inequality (why?). To do that, move the left hand side to the right hand side. You want to show the function on the right hand side now is always greater than 0. When $$y=\mu$$, the function is positive and the limit as y goes to infinity is 0. Take the derivative with respect to y. Show the derivative is positive at $$y=\mu$$ and there is one $$y>\mu$$ where the derivative is 0. These four facts will prove that the function on the RHS is always positive for $$y>\mu$$.
d is false. Take $$y=\mu=0$$ and $$\sigma^2=\frac{8}{\pi}$$. LHS is 1/2, RHS is 1/4.
• The derivative, $(\frac{1}{\sqrt{2\pi}\sigma}-\frac{y-\mu}{\sigma^2})\exp(-\frac{(y-\mu)^2}{2\sigma^2})$, may not be positive at $y=0$. Jan 22, 2021 at 18:10
• Right, I meant $y=\mu$. I was thinking $\mu=0$ without loss of generality, but I edited my answer now to say it correctly. Jan 22, 2021 at 19:01