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How can I find unbiased estimator for $\theta$ for r.v. with p.d.f. $f_X(x)=e^{-\theta x}$, $x>0$, $θ>0$. Let $(X_1,\ldots,X_n)$ be a s.r.s. of such r.v.

I think I need to find expectation of the random variable to get the estimator $\hat \theta$ and to check whether it unbiased or not, I need to know the pdf $\sum_{i=1}^n X_i$ , and find the expectation of $\hat \theta$ but how can I find it?

I'm trying to integrate to get expectation, and got $E[x]=\int\limits_{-\infty}^{\infty}f_X(x)x= \frac{1}{\theta^2}$

but I also see some explanation about this is being $X\sim \mathrm{Exp}(1/\theta)$ exponential distribution with $E[x]=\frac{1}{\theta}$ , but I don't know which one is right? I know that exponential distribution p.d.f is

$$X\sim \mathrm{Exp}(\lambda):f(x;\lambda)=\lambda e^{-\lambda x}$$

StubbornAtom
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vedss
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    If $X$ has pdf $f_X(x)=\theta e^{-\theta x}1_{x>0}$ (yours is not a pdf), then $E[X]=\int_0^\infty x\theta e^{-\theta x},dx=\frac1{\theta}$. I believe what you are asking for is this. – StubbornAtom Nov 02 '19 at 11:25
  • @StubbornAtom thankyou so much! i see this website https://bookdown.org/egarpor/inference/point-est.html#point-est-unbiased it said that in ex 3.4 $ f_{X}(x)=e^{-\theta x}, \ x>0, \ \theta>0.$ and $X\sim \mathrm{Exp}(1/\theta)$ why the parameter is $\frac{1}{\theta}$? is the p.d.f right too? – vedss Nov 03 '19 at 08:11
  • They made a mistake in the pdf, as you should be able to check whether $f_X$ integrates to one or not. As for why they write $X\sim \text{Exp}(1/\theta)$, it is just notation (there are two parametrizations of the one-parameter exponential distribution). You could have called the same distribution $\text{Exp}(\theta)$, it does not matter. – StubbornAtom Nov 03 '19 at 08:34

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