# How do we obtain an estimate for $\textbf{P}(X\geq 1)$ where $X\sim\text{Exp}(\lambda)$?

Consider a simple sample $$X_{1},X_{2},\ldots,X_{n}$$ whose distribution is given by $$X\sim Exp(\lambda)$$.

(a) Determine an estimator for $$\lambda$$ according to the method of moments.

(b) Determine another estimator for $$\lambda$$ different from the previous one.

(c) Determine an estimate of $$\textbf{P}(X\geq 1)$$ in accordance to the method of moments.

MY ATTEMPT

As to the first case, we have \begin{align*} \frac{1}{\lambda} = \textbf{E}(X) \Rightarrow \hat{\lambda} = \frac{n}{\displaystyle\sum_{k=1}^{n}X_{k}} \end{align*}

As to the second case, we have \begin{align*} \frac{1}{\lambda^{2}} = \textbf{Var}(X) = \textbf{E}(X^{2}) - \textbf{E}(X)^{2} \Rightarrow \hat{\lambda} = \left[\frac{1}{n}\sum_{k=1}^{n}X^{2}_{k} - \left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)^{2}\right]^{-1/2} \end{align*}

EDIT

In the third case, we have \begin{align*} \textbf{P}(X\geq 1) = 1 - \textbf{P}(X < 1) = 1 - \exp(-1\times\lambda) = 1 - \exp(-\lambda) \end{align*}

Now it suffices to substitute $$\lambda$$ for any of the above expressions.

Could someone double-check my reasoning? Thanks in advance!

• Exponential distribution is continuous, but you're treating it as discrete.... is there a typo? – Lee David Chung Lin Apr 8 at 2:00
• Indeed, you are right. I was thinking about the poisson distribution when I wrote it. Thanks for the observation. I will edit it. – user1337 Apr 8 at 2:08
• What have you tried? You know, don't you, that this is not a site for us to do your homework for you? – David G. Stork Apr 8 at 2:22
• Plug the estimator in $(a)$ into $e^{-\lambda}$. – d.k.o. Apr 8 at 2:27
• Sorry, David. I know it. I just forgot to mention my attempts. – user1337 Apr 8 at 2:38