exponential distribution of an exponential variable

Question

Given a random variable Y has an exponential distribution with mean 2. Let $h(y) = \exp(-Y)$. How could I find $P[h(y) \leq 0.35]$?

For $Y$, I think $f(y, \beta) = \frac{1}{\beta} \exp(-y/\beta) = \frac{1}{2} \exp(-2y)$, but am not sure how to find $P[h(Y)\leq 0.35]$.

Could anyone please explain how to go about this question?

Answer 1

Let $Y \sim {\cal E}(\lambda)$ with $\lambda>0$. Then $E[Y] = \lambda^{-1} = \beta$.

• The density function is $f(y, \beta) = \exp(-y/\beta)/\beta$ for every $y \ge 0$ and $\beta > 0$.
• The distribution function is $F(y,\beta) = 1-\exp(-y/\beta)$ for every $y \ge 0$ and $\beta > 0$.

In this question, $\beta = 2$ and $\lambda = 1/2$, hence \begin{align} \Bbb{P}(\exp(-Y) \le 0.35) &= \Bbb{P}(Y \ge -\log0.35) \\ &= \exp(-(-\log0.35)/2) \\ &= \exp(\log\sqrt{0.35}) \\ &= \sqrt{0.35} \\ &= \sqrt{\frac{7}{20}} \end{align}