Let X have the Pareto distribution with...

Let X have the Pareto distribution with pdf egin{equation*} f(x;k, heta) = egin{cases} frac{k heta^k}{x^{k+1}} & x ge heta \ 0 & x < heta, end{cases} end{equation*} where $k, heta$ are fixed positive numbers (parameters). (a) If $k > 1$, find $E(X)$ (which shall be a function of $k$ and $ heta$). (b) If $k = 1$, what can you say about $E(X)$? (c) If $k > 2$, find $V(X)$. (d) If $k = 2$, what can you say about $V(X)$?

Transcript

00:01 So in this question we're told that x has a per rata distribution with pdf, f of x with parameters k and theta, is equal to k theta to the k divided by x to k plus 1.

00:18 Sorry, that's kind of hard to read.

00:22 X to the power of k plus 1 for x greater than or equal to theta and zero for x less than theta.

00:29 So first of all, let's find the expected value of x.

00:35 Well, the expected value of x is the integral from theta to infinity of x times the distribution here.

00:46 So multiplying x into that distribution, we're going to get k, theta to the k divided by x to the k, which we integrate with respect to x.

00:55 Now we're told that k is greater than 1.

01:03 So we have for k greater than 1, this is going to be k theta to the k, 1 over x to the k minus 1.

01:16 And we're going to have to divide by a factor of 1 minus k.

01:21 Because when we differentiate this, we'll get a factor of minus k minus 1, which is 1 minus k.

01:28 And we evaluate that between theta and infinity.

01:35 Now since k is greater than one, this is a truly negative power of x, which means that at infinity, this is going to go to zero.

01:44 So at infinity, we don't get anything.

01:46 At theta, we take a minus sign, so this becomes k over 1 minus k, and we have theta to the k over theta to the k minus 1, which is just theta.

01:57 So that's our expected value of x for k greater than 1.

02:07 So this is the answer to part a.

02:10 Now for part b, what about the expected value of x when k is equal to 1? well, let's go back to our integral.

02:20 We're integrating from theta to infinity times theta over x, dx.

02:34 So this is going to be theta log x, evaluated between theta and infinity.

02:41 So this is theta log infinity minus theta.

02:45 Log theta, but log of infinity is undefined since the limit as x goes to infinity log x is infinite.

03:06 So this expectation value does not exist.

03:12 So the expectation value of x with k equals 1 does not exist.

03:20 To be formal, more informally we could say informally, it's equal to infinity...

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