00:02
Okay, this question gives us this moment generating function, and it wants us to find the original probability distribution, along with the expected value and standard deviation.
00:16
To start, we're going to find the original probability distribution.
00:21
And we can do that by remembering that in order to get this moment generating function, you would have had to add the probabilities times e to the corresponding x values.
00:36
With a t attached.
00:40
And we can see here that if we look at this moment generating function, as long as we just insert an e to the zero here, we see that this follows this format exactly, because we have 0 .2, which is a probability, times e to the 0, plus 0 .3 times e to the 1t, plus 0 .5e to the 3t.
01:05
So we have a probability distribution that can take on three values.
01:11
So just drawing a little table, here for x and p of x or sorry this should be 0 1 and 3 because those are the only 3 exponents we see for our moment generating function so p of 0 .0 it tells us as 0 .2 based on this p of 1 it tells us as 0 .3 based on this and p of 3 is 0 .5 based on this and we see that this is indeed a valid probability distribution because it adds up to one and has probabilities in the correct ranges.
02:00
So now we want to find the expected value and variance.
02:06
So we'll have to find at least two derivatives.
02:11
So our first derivative, which is going to relate to our expected value, well, just taking a t derivative, we get 0 plus 0 .3e to the t, plus, and then for the 3t term, we just have to use the chain rule to get 1 .5, e to the 3t.
02:35
And the expected value is just this moment generating function's first derivative at 0.
02:46
So we just get 0 plus 0 .3 plus 1 .5 for an expected value of 1 .8...