# Thread: estimator of theta

1. ## estimator of theta

If X1, X2.......,Xn constitute a random sample of size n from an exponential population, show that "X bar" is a sufficient estimator of the parameter theta

Can somebody please, please help me with this? It is giving me a lot of trouble

2. Originally Posted by Moo
Thank you for the hint, but I already knew that factorization theorem was involved but I just can't seem to solve it. I was hoping someone could give me a more step by step answer because I am really trying but have no outside help for this course and the professor is hardly ever available to assist.

3. Just write your likelihood function which is the joint density....

$f(x_1,\cdots,x_n)={1\over \theta^n}e^{-\sum_{i=1}^nx_i/\theta}$

where I left out the fact that all the x's are positive, that usually just confuses people.
meaning........

$f(x_1,\cdots,x_n)={1\over \theta^n}e^{-\sum_{i=1}^nx_i/\theta}I(x_1>0)\cdots I(x_n>0)$

This shows that the SUM is suff for theta, but so is twice the sum....
so the sum divided by n is suff....

$f(x_1,\cdots,x_n)={1\over \theta^n}e^{-n\bar X/\theta}I(x_1>0)\cdots I(x_n>0)$

shows that the sample mean is suff for theta.

4. ## please verify my answer. Need help fast

Not sure if I have this right but I want to make sure the way I am doing it is correct. Thanks

If X1,.......,Xn make up a random sample of size n from an exponential population, show that 'X bar' is a sufficient estimator of the parameter theta

Not sure how to write symbols, so I am going to let
Theta = Z
Summation = S
e^ = exp(...)

So,

Z^-n * exp(-S(Xi)/Z)

take LN

= -n LN(Z) - S(Xi)/Z

Take derivate w.r.t. Z

= -n/Z + S(Xi)/Z^2

Set = 0 and solve for Z

0 = -nZ + S(Xi)
nZ = S(Xi)
Z = S(Xi)/n

Is this right?

### if x1 constitute a random sample from a exponential population

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