00:01
Okay, so there's a lot of questions here.
00:04
Now we can only do one question per video, and i think we've answered questions one and two in a different video, so i'm going to do question three.
00:09
If you want the other answering as well, if you could just please upload them all as separate questions, then that would be really helpful so that we can give the appropriate amount of time to each one rather than rushing through them.
00:20
So for question three, for part a, we want the value of alpha such that f of x is a probability density function.
00:27
And we know that in order for it to be a pdf, it needs to integrate to 1 over the allowed values of x, which here is 1 to infinity.
00:36
So this is the integral from 1 to infinity of alpha over x to the k.
00:42
And that integral is just equal to, so this is, we could write as alpha x to the minus k.
00:50
So we could do this integral as minus 1 over k minus 1 times alpha x to the minus k minus 1.
01:03
Now, evaluating at infinity, this vanishes because you've got x to the negative power of infinity, so that's 1 over x to the infinity or 1 over infinity kind of, which gives us 0.
01:19
And we've got, oh, sorry, just 1 over infinity to the k minus 1, so that's going to give a 0.
01:24
And then when x equals 1, so we're going to get 0 minus 1 over k minus 1 times alpha, and then times 1 to some power, which just gives us 1.
01:32
So we find this is alpha over k minus 1, and we know that has to equal 1, therefore alpha has to equal k minus 1.
01:42
The expected value then is just given by the integral of the pdf times x...