Hey everyone, in this video I want to do this statistics and probability problem. This is a
second example of how to calculate a methods of moment estimator when you're given f of x, which
is a pdf. I already have a video on calculating methods of moments on my channel--I actually
have two videos. This is just going to be another example with another pdf. And so let's go ahead
and get started with solving this problem. If the problem does not tell you which moment to use,
the simplest moment to use is just the first moment. You're going to want to calculate the
expected value of x, which is equal to the mean, and you're given the pdf. If you're given the pdf,
one thing you can do is calculate the expected value by hand and the expected value is going to
be equal to the integral from negative infinity to infinity of x times f of x, which is the pdf,
dx. Now this problem was given to me as a homework problem and I would argue that we are not expected
to calculate the expected value of this pdf. I think that for this problem what we're expected
to do is to be able to recognize what pdf or what probability distribution this is, given the pdf.
And then I think that we are more likely expected to find or look up the expected value of the pdf
and then go from there and solve for the method of moments estimator using the expected value that we
look up. And I'll explain in a second when I show you what the expected value looks like for this
pdf, it will kind of make sense why I think we're not supposed to find it by hand. For the pdf that
we're given, I know that when we have a PDF that's proportional to basically x times e to the x,
that's known as a gamma distribution and that has its own expected value. In this particular case,
we have a slightly different pdf or different function because we have something proportional
to x times e to the x to the delta. This pdf is actually a more generalized version of a gamma
distribution, so it's a generalized gamma known as the Weibull distribution. And so
it's considered a generalized form of the gamma distribution because if delta is equal to one,
then you would have a gamma distribution but delta can be equal to something that's not
equal to one and so it would be slightly a more complicated function. And so if I was doing this
as a homework problem and I had access to the internet, once I'm able to identify the given
pdf for the question I would then look up the probability distribution for the given pdf on
wikipedia. I just said that we were working with the Weibull distribution. When I look that up,
the probability distributions on wikipedia on the right hand side have these nice info boxes where
you have different properties of the probability distribution given to you. You have the pdf, cdf,
mean, median, mode, and other characteristics as well. Now for our particular problem, we're
not using the standard parameterization. Instead, we're given it in the form called the alternative
parameterization and so this is the pdf that we were given in the problem. And if I look it up,
the expected value or the mean of the Weibull distribution is this function here. This is a
function or the result that I'm going to use as the expected value to solve the rest of the
problem. All right so going back to the problem that we were given, the way you calculate a
method of moments estimator is we're going to find the expected value, that's equal to some
value. Then we're going to estimate the sample mean by plugging it into the expected value or
the population mean. And then we're going to solve for the parameter that we're looking for,
which in this case is beta. On the right hand side, we're going to have something, something,
something, beta, then we're going to estimate it and solve for beta using the sample mean instead
of the expected value. Using the expected value that we got from wikipedia, the expected value
for the alternative parameterization of the Weibull distribution is equal to 1 over beta
to the 1 over delta times the gamma function of 1 plus 1 over delta. And since this is kind
of like a weird looking function, this is why I think that we were not expected to calculate the
expected value by hand because I don't think we were expected to be able to figure out that the
integral is equal to some kind of gamma function. I just think we were meant to look that up. And so
following the steps for calculating the MOM, we're just going to plug into the sample mean and now we
want to estimate and solve for beta hat. That's going to be equal to the sample mean of 1 over
beta hat to the 1 over delta times the gamma function of 1 plus 1 over delta. I'm going to
multiply both sides by beta to the 1 over delta and divide by the sample mean. That's going to
give me beta hat 1 over delta is equal to 1 over the sample mean times the gamma function of 1 plus
1 over delta and then I'm just going to raise both sides by the exponent delta. And that's
going to give us our final solution. Beta hat, the methods of moments estimator for beta, is equal to
gamma 1 plus 1 over delta over the sample mean to the delta. And we are done with this problem.
*chiptune music*