Is $\exp(-2/X_1)$ an unbiased estimator of $\exp(-2/\theta)$? If $X_1, ..., X_n$ are i.i.d. with pdf
$$f(x\mid\theta) = \frac{1}{\theta}e^{-x/\theta}\quad x&gt;0,\, \theta&gt;0$$
I would like to estimate $P (X_1 &gt; 2) = e^{-2/\theta}$. I think $e^{-2/X_1}$ is a biased estimator, but I don't know how to prove. I tried to prove by definition,
$$ E(e^{-2/X_1}) = \int_{0}^{\infty} e^{-2/x}\frac{1}{\theta}e^{-x/\theta}dx$$
but the integral is very complex to evaluate. Are there any other ways to prove it? Thanks in advance!
If I understand your question, I believe you can directly compute $P\left(X_{1} \gt 2\right)$ from:
$$
P\left(X_{1} \gt 2\right)= \int_{2}^{\infty}\frac{e^{-x/\theta}}{\theta}\mathrm{d}x
$$
$$
P\left(X_{1} \gt 2\right)= -\frac{1}{\theta} \theta \left[e^{-x/\theta}\right]_{2}^{\infty}=e^{-2/\theta}
$$
I hope this helps.
If I understand your question, I believe you can directly compute $P\left(X_{1} \gt 2\right)$ from:
$$
P\left(X_{1} \gt 2\right)= \int_{2}^{\infty}\frac{e^{-x/\theta}}{\theta}\mathrm{d}x
$$
$$
P\left(X_{1} \gt 2\right)= -\frac{1}{\theta} \theta \left[e^{-x/\theta}\right]_{2}^{\infty}=e^{-2/\theta}
$$
I hope this helps.