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MLE exponential

Hi guys, I'm stuck on one problem, regarding likelihood functions. I found the MLE of the exponential:

y=(λ)exp[(λ)x] y = (\lambda)exp[-(\lambda)x]

And the MLE was found to be:

λ ^=1/x \lambda\hat\ = 1/x

But I'm stuck on answering the question about whether it's unbiased and consistent. I know that for an estimator to be consistent, its variance approaches 0 as n increases, and that the biased formula is:

Bias(X)^=E(X)^X Bias(X\hat) = E(X\hat) - X

So I apply it to my estimator and get this:

Bias[λ]^=E(1/x)x Bias[\lambda\hat] = E(1/x) - x

So, where do I go from here? My lecturer told me that it can "easily be found by integration" but how? Any help would be appreciated :smile:
I'm not sure how you go about integrating the bais, but E(1/x) - x , seems like an exponetial and x integration
Reply 2
Original post by devangdave
I'm not sure how you go about integrating the bais, but E(1/x) - x , seems like an exponetial and x integration


Yeah, I'm not exactly sure about how to do E(1/x). My lecturer claims it's easy, but I don't see how to do it.
is that meant to be e^(1/x)?
(What's in the **stars** is incorrect)

Spoiler

(edited 9 years ago)
Reply 5
Original post by rayquaza17
E[1/x]=1/E[x]

Do you know how to calculate E[x] for an exponential distribution?

Shouldn't the bias be:
Bias(λ^)=E[λ^]λ=E[1x]λ=1E[x]λBias( \hat{\lambda})=E[\hat{\lambda}]-\lambda=E[\frac{1}{x}]-\lambda=\frac{1}{E[x]}-\lambda


Oh, thank you :smile:

Now I can use the mean proof for the exponential to arrive at the answer :smile:
Reply 6
Original post by devangdave
is that meant to be e^(1/x)?


No, it's E(1/x) because it's an expected value.
Original post by SecretDuck
Oh, thank you :smile:

Now I can use the mean proof for the exponential to arrive at the answer :smile:


No problems. :biggrin:
Original post by SecretDuck
.


I'm really sorry about this but:

I've had a think about it, and I don't think E[1/x]=1/E[x] so I think what I've said above is incorrect.

I've had a look on google, and it seems as if you need to use the gamma distribution to find the required bias. I do not think this can easily be found by integration at all?? TBH I think you should speak to your lecturer again because unless I'm missing something, this is a really difficult question!

If you're meant to find the asymptotic bias though, the answer is 0 because MLEs are always asymptotically unbiased.

Sorry, I feel really bad telling you the wrong thing. :frown:
Reply 9
Original post by rayquaza17
I'm really sorry about this but:

I've had a think about it, and I don't think E[1/x]=1/E[x] so I think what I've said above is incorrect.

I've had a look on google, and it seems as if you need to use the gamma distribution to find the required bias. I do not think this can easily be found by integration at all?? TBH I think you should speak to your lecturer again because unless I'm missing something, this is a really difficult question!

If you're meant to find the asymptotic bias though, the answer is 0 because MLEs are always asymptotically unbiased.

Sorry, I feel really bad telling you the wrong thing. :frown:


Oh right, thanks anyway :smile:

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