A relationship about normal distribution

Question

Given two random variables, $X\sim N(\mu_{1},\sigma_{1})$ and $Y\sim N(\mu_{2}, \sigma_{2})$ where $\mu_{1},\mu_{2}>0$.

If $\sigma_{1}<\sigma_{2}$ and $a>1$, Can we conclude the following relationship? $$P(X>E[X]a)<P(Y>E[Y]a).$$

Answer 1

Write instead $$P((X-\mu_1)/\sigma_1 > \mu_1(a - 1)/\sigma_1) = 1 - \Phi(\mu_1(a-1)/\sigma_1)$$ and similarly for $Y$, where $\Phi$ is the normal cdf. Hence the statement is equivalent to $$ \Phi(\mu_1(a-1)/\sigma_1) > \Phi(\mu_2(a-1)/\sigma_2), $$ which due to monotonicity is equivalent to $$ \frac{\mu_1(a-1)}{\sigma_1} > \frac{\mu_2(a-1)}{\sigma_2}. $$ To this end, since $a > 1$, $$ \frac{(a-1)(\mu_1\sigma_2 - \mu_2\sigma_1)}{\sigma_1\sigma_2} > 0 \iff \mu_1\sigma_2 > \mu_2\sigma_1. $$ Notice that $\sigma_1 < \sigma_2$ is not needed.