Maximum liklihood estimators again help

[semidevil]

I dotn know, I'm still lost on this whole MLE thing...but here is my attempt at some problems...please critique(just the concept is still bugging me).

find the MLE:

$$p_x(k;\theta) = \theta^k (1-\theta)^{1-k}, . k = 0, 1, 0 < \theta < 1$$.

so here is what I did.

$$L(\theta) = \theta^{nk} ( 1- \theta)^{{\sum_1^n{n - k}}$$

$$ln L(\theta) = nkln\theta + \sum_1^k 1-k * ln(1-\theta)$$

now, take derivative

$$nk/\theta + \sum_1^k/{1-\theta}$$.

first of all, in geting the formula, is this right? I know I will need to leave it in terms of theta, but I don't know if even this is right??

 

[Nate810]

now, to find maximum, set the derivative to 0, so 0 = nk/\theta + \sum_1^k/{1-\theta} and then solve for theta\theta = nk/{\sum_1^k 1-k} so MLE = \theta = nk/{\sum_1^k 1-k} Please let me know if this is correct, or if I am just completely off track!

 

[wais]

Hi there,

Thank you for sharing your attempt at finding the MLE for this problem. Overall, your approach is correct, but there are a few minor mistakes that need to be addressed.

First, when taking the log likelihood, you need to use the entire expression for p_x(k; \theta), not just the exponent. So the correct expression for ln L(\theta) should be:

ln L(\theta) = kln\theta + (n-k)ln(1-\theta)

Also, when taking the derivative, you need to use the chain rule. So the correct derivative would be:

d/d\theta ln L(\theta) = (k/\theta) + ((n-k)/(1-\theta))(-1)

= k/\theta - (n-k)/(1-\theta)

= (k-n\theta)/(1-\theta)

Finally, to find the MLE, you need to set this derivative equal to 0 and solve for \theta. So you would have:

(k-n\theta)/(1-\theta) = 0

k-n\theta = 0

n\theta = k

\theta = k/n

Therefore, the MLE for this problem is \theta = k/n.

I hope this helps clarify the concept of MLE for you. Just remember to use the entire expression for p_x(k; \theta) when taking the log likelihood, and to use the chain rule when taking derivatives. Keep practicing and you'll get the hang of it!