Hey guys, in this video I'm going to go over these probability problems that are based on
a multivariate normal distribution. In this case, we have a random vector X, which is three random
variable--a vector of three random variables that are normally distributed and we know it's three
because of the three here. And we are given the two parameters. We're given the matrix of
the means for X1, X2, X3 and we are given the covariance matrix of X. The covariance matrix
has the variances on the main diagonal, so this is the variance of X1, variance of X2, variance
of X3. And then the off diagonal portions have the covariances. This value for example is the
covariance between X1, because it's the first row, and X2, because it's the second column. And we can
use that to find these different probabilities. Let's go ahead and get started with the first
one. This is just finding the probability that X1 is greater than 6. The idea with
this is we're going to write out what the normal distribution is for each of these transformations,
or parts of the given distribution. And then we're going to convert these to a Z-score and
from the Z-score we're going to use a Z table to find the probabilities. And you can find a Z table
online or I'll put a link in the description box to a PDF. Okay so we're going to find the
mu and the variance for X1 in order to convert it to a Z-score. Looking at the mu vector,
mu for X1 is just 5 and then the variance for X1 is just 4. We know that X is normally distributed
with mu or mean 5 and variance 4. The formula for a Z-score is X minus mu over the standard
deviation, so in that case this is going to be X, which is what we are looking for, minus mu,
which is 5, over the square root of 4, which is 2. This is going to give us 0.5. The probability
of Z being greater than 0.5. On the table that I'm using, we're looking at lower than values,
so this is equal to the probability Z is less than -0.5 because of symmetry. And the probability
that you get if you look this up on the table is 0.3085 and that's the answer for part A. Part A
is done. Part B is a linear transformation of two random normally distributed variables. There's a
theorem that says that any linear combination of normally distributed variables is also normally
distributed, so 5 X2 + 4 X3, we can say that we know that that is normally distributed because we
know that X2 and X3 are normally distributed. And again we have to find mu and sigma. Let's go ahead
and do that. For the mu, you just have to get the expected value of what we're asked to find. The
expected value is pretty straightforward, we just do the expected value of 5 X2, which is equal to
5, times the expected value of 2, which is equal to mu2 plus 4 mu3. And if we look for X2 and X3,
that's 3 and 7, so that's going to be 5 times 3 plus 4 times 7 and that's going to be 43. Mu is
going to be 43. There is more than one way to find the variance. I'll do one way for part B
and another way for a part D. If it's just two variables, I think it's maybe easier to do just
a binomial expansion, so 5 X2 plus 4 X3, multiply that by itself because you're finding the variance
and then if you multiply this out, that's 5 X2 squared plus 40 times the covariance of X2,
X3 plus 16 X3 squared. This is equal to 25 of the variance of X2 plus 40 covariance X2,
X3 plus 16 times the variance of X3. And then we are given the covariance matrix,
so we can find the covariance of 2 and 3, and we're given the variances as well obviously.
That's going to be 25 times the variance of X2. That's going to just be 4 from looking
at the covariance matrix. That's that. Plus 40 times the covariance of X2 and X3 so 2, 3,
that's 2. And then plus 16 times the variance of 3 and that will be 9. And if we add and multiply,
that's going to be 324 for the variance. So 5 X2 plus 4 X3 is normally distributed with 43
as the mean and 324 as the variance. We are asked to find the probability that that is greater than
70. When we find the Z-score, we're going to do 70 minus 43 over the square root of 324. That
gives us 1.5. The probability Z is greater than 1.5 is equal to the probability that Z
is less than -1.5 and that is equal to .0668 according to the table. And we are done with
that problem. Now let's just do the last one and this is looking at a linear transformation with
the three variables. I'm going to find variance in a different way. Let me just erase this so I
don't have to scroll down too much. D is asking to find the probability 4 X1 minus 3 X2 plus 5
X3. We want to find out again the mu and the variance. Doing the same thing as last time,
this is going to give us 46 and then another way of finding the variance is using matrix
multiplication and I think with three variables or more it's easier to use matrix multiplication.
This is going to be A times the covariance matrix X times A transpose, and this is matrix
multiplication. So A is the coefficient vector in this case. That's going to be 4 minus 3 5 times
the covariance matrix, which we are given, and it is (4 -1 0; -1 4 2; 0 2 9) times A transpose,
so that's 4 -3 5, so if we do that that's going to give us one single value at the end but the
steps are just doing matrix multiplication left to right. That's 16 plus 3, -4 minus 12 plus 10,
-6 plus 45 and then that's going to be times 4 -3 5. And the final answer for the variance is
289. So we have that this is normally distributed with 46 as the mean and 289 as the variance and
again like all the other problems we're going to transform into a Z-score that equals 2.0. And
we are asked to find the probability that it's less than, so we don't have to use symmetries,
so the probability that Z is less than 2 is 0.9772 and we are done with this problem.
*chiptune music*