probability of exponential distribution question

Question

Suppose $X_n$ follows Exponential distribution with parameter $\alpha$. Find P($X_n$ < log(n)*$\epsilon$). I got my answer equals 1 - $\frac{1}{n^(\alpha\epsilon)}$, which is quite different from the answer. I wonder am I right?

Answer 1

If your density is

$$f_{x_n}(t)=\alpha e^{-\alpha t}$$

your CDF is

$$F_{X_n}(t)=1-e^{-\alpha t}$$

thus

$$F_{X_n}(\epsilon \log n)=1-e^{-\alpha \epsilon \log n}=1-\left[ e^{\log n}\right]^{-\alpha\epsilon}=1-\frac{1}{n^{\alpha\epsilon}}$$