X~Poi(lambda) and Y~Poi(2*lambda). X and Y are independent. I am to find MLE estimator of lambda. I am forming the joint density and extracting marginal for X. Is this necessary? I know f(x).
Poisson Example
P(X = x) =
lambda(x) e(- lambda)/x!
For X1;X2; : : : ;Xn iid Poisson random variables will have a joint frequency function that is a product of the marginal frequency functions, the log likelihood will thus be:
l(lambda) =sum{i=1}{n}(Xiloglambda- lambda- lambda logXi!)
= loglambda sum{i=1}{n} Xi -nlambda- lambda sum{i=1}{n} logXi!
We need to find the maximum by finding the derivative:
l'(lambda) =sum{i=1}{n} xi -n =o
which implies that the estimate should be
lambda^= X bar.
The mle of the Poisson pmf is meaningless. However, the mle of lambda is the sample mean of the distribution of X. The mle of lambda is a half the sample mean of the distribution of Y. If we must combine the distributions the lambda may be estimated by (sample mean of X + (sample mean of Y)/2)/2.
Ette: How do we combine the distributions? Will (sample mean of X + (sample mean of Y)/2)/2. be an MLE of lambda? The question of Mary is not correct?
I am embarrassed to say that I got some of my log expressions mixed up. That completely affected the answer. The derivative is taken with respect to lam not with respect to x and y which are now constants! The correct procedure is given below.
I assume that X and Y are independent. Then likelihood function is the joint pdf which is{(lam^x)exp(-lam)/x!} {((2* lam)^y)exp(-2*lam)/y!}. It is a function of one variable lam not two since population mean of y is twice that of x.
Compute the log of the likelihood function and then take the derivative of the result. Set equal to zero and solve for lam. You get lam=(x+y)/3.
Such fun. Cheers!
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