# MLE (Statistics)

1. Jan 22, 2008

### J Flanders

This is my question: Find the Maximum Likelihood Estimator for
f(y / x) = (x + 1)y^x, 0 < y < 1 and x > -1 OR 0, elsewhere.

I think this is how you get started, but I get confused. I'm not sure how to continue.
The likelihood function defined as the joint density of Y1, Y2, ...., Yn evaluated at y1, y2, ..., yn is given by
L = product from i = 1 to n of (x + 1)(yi^x) = (x + 1)^n * product from i = 1 to n of (yi^x).
I'm sorry for the notation. Any help is obviously appreciated.

2. Jan 22, 2008

### EnumaElish

Your question can be restated as:

Find the MLE of x given:

f(y | x) = (x + 1)y^x, 0 < y < 1 and x > -1
f(y | x) = 0 elsewhere.

I am assuming that (i) f is the pdf of "y given x" and (ii) the y's are independent.

Then, the simplest procedure would be to take the log of L(x) = $\prod_{i = 1}^n (x + 1)y_i^x$ and maximize it with respect to x.

Last edited: Jan 22, 2008
3. Jan 22, 2008

### J Flanders

So I get:

ln[((x+1)^n)*(y1^x)*(y2^x)*....*(yn^x)], but I don't see how you maximize this.

I would imagine you take the derivative and set it equal to zero, but I cannot solve for x. What is the maximum value for x?

Thanks for the help from before.

4. Jan 27, 2008

### J Flanders

OK, I just want to check this.

I get:
ln[((x+1)^n)*(y1^x)*(y2^x)*....*(yn^x)] =
nln(x+1) + x[ln(y1) + ln(y2) + ... + ln(yn)

I take the derivative with respect to x and set it equal to zero:
n/(x+1) + ln(y1) + ln(y2) + .... + y(n)
and with algebra this implies

x (the estimator) = [-n - ln(y1) - ln(y2) - ... - ln(yn)] / [ln(y1) + ... + ln(yn)

Does this make sense? Thanks again for your help; it is appreciated.