00:01
Now here on this problem, we are told that we have a random variable y with a moment generating function m of t.
00:09
We're also told that we have u that's equal to a, y, plus b.
00:15
And we'd like to find its moment generating function and then it's expected value.
00:21
Now, the moment generating function for you is equal to the moment generating function of ay plus b, which means that this is the expected value of e to the ay plus b.
00:45
T.
00:47
E to the a .y plus b .t.
00:49
I use our properties of exponents here.
00:51
This is equal to the expected value of e to the ayt times e to the bt.
01:04
Now since e to the bt is just a constant, this means this is e to the bt times the expected value of.
01:12
Of e to the ayt.
01:18
Now e to the ayt is just the moment generating function of y evaluated at a .t.
01:26
And so the moment generating function of u is equal to e to the bt as a moment generating function of y at a .t.
01:34
And that's what we wanted to show here.
01:37
Now we want to use this to find the variance and the expected value for you.
01:45
Now the expected value of you is equal to the moment generating function.
01:49
Of u's derivative at zero.
01:54
First, let's find our derivative here.
01:58
Using the product rule, this is equal to b, e to the b, t, times m of a, t, plus a, m prime of a t, e to the b t.
02:17
And so m, u prime of zero is b, e to the zero, m of zero, plus a, m prime of zero, plus a, m prime of zero, e to the zero.
02:37
E to the 0 and m of 0 are both 1, and so that term there just becomes b.
02:42
M prime of 0 is the expected value of y.
02:47
And so this becomes a times mu, because mu is the expected value of y.
02:55
And so the expected value of u is b plus a mu.
03:01
That's the expected value of mu.
03:04
Now for the variance, we need the second derivative of you.
03:09
And so we take the derivative of the first derivative...