Find MLE for f(y/x) = (x + 1)y^x, 0 < y < 1 and x > -1

In summary, the Maximum Likelihood Estimator for f(y / x) = (x + 1)y^x, 0 < y < 1 and x > -1 OR 0, elsewhere is obtained by taking the log of the likelihood function and maximizing it with respect to x. This results in x = [-n - ln(y1) - ln(y2) - ... - ln(yn)] / [ln(y1) + ... + ln(yn)].
  • #1
J Flanders
6
0
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.
 
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  • #2
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) = [itex]\prod_{i = 1}^n (x + 1)y_i^x[/itex] and maximize it with respect to x.
 
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  • #3
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
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.
 

FAQ: Find MLE for f(y/x) = (x + 1)y^x, 0 < y < 1 and x > -1

What is the maximum likelihood estimation (MLE) for f(y/x)?

The MLE for f(y/x) is the value of y that maximizes the likelihood function for a given x. In this case, it is the value of y that makes the expression (x + 1)y^x the largest.

How do you find the MLE for f(y/x)?

To find the MLE for f(y/x), you would first take the derivative of the likelihood function with respect to y, set it equal to 0, and solve for y. This value of y will give you the MLE for f(y/x).

What are the restrictions on y and x for this problem?

The restrictions on y are that it must be greater than 0 and less than 1. The restriction on x is that it must be greater than -1.

Is the MLE for f(y/x) unique?

In this case, the MLE for f(y/x) is not unique. There are multiple values of y that can maximize the likelihood function for a given x. However, there is one y value that will give the absolute maximum likelihood.

Can the MLE for f(y/x) be negative?

No, the MLE for f(y/x) cannot be negative. Since y is raised to the power of x, it would result in a negative value if y is negative. However, the restrictions on y do not allow for negative values.

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