2. (10 points) Consider a random process process X⃗∼ N3(μ⃗,...

2. (10 points) Consider a random process process $\vec{X} \sim N_3(\vec{\mu}, \Sigma)$ whose mean vector $\vec{\mu} = \begin{bmatrix} 70\\50\\15 \end{bmatrix}$ and Covariance matrix $\Sigma = \begin{bmatrix} 8 & 0 & 0\\0 & 4 & 0\\0 & 0 & 1 \end{bmatrix}$ obtain the trivariate normal distribution of $\vec{X}$

Transcript

00:01 Here in this question we are given that the random vector, random vector which is vector x as x1, x2, x3, x4, and x pi, with mean, with mean vector and variance, covariance matrix, variance, variance, o variance matrix as mu is given here, r variant mean is 1, 2, minus 1, 3 and 0.

01:06 And value of cimation is given here as in the shape of matrix 8 minus 2, 1 minus 1, 0.

01:17 Second row is minus 2 6 2 minus 2 0 then we have 1 2 12 12 2 minus 2 and now next row is minus 1 minus 2 2 8 0 and our last row is here as 0 0 minus 2 minus 2 0 and 4 in the matrix.

01:54 Respectively and now we can find our value for partition x as value of partition x as here x1, x2, x3, x4, x5 and our partition is here between x2 and x3 then this will be equal to x2, x3 divided by.

02:30 Add x2.

02:35 So now let's take a is as matrix a is as 1 minus 1 1 and 1 and here next b is 1 1 1 and 2nd row will be 1 1 and minus 2 now to solve part a here what we need to do we will find our will be summation of a x of 1 that will be as a, e x similar to 1 and this will be as equal to a e for x1 x2 which is a of e x1 e x2 will be equal to a of 1 and 2 and therefore we can put our values here as a is 1 minus 1 1 1 and 1 x1 x1 x1 and x2 are 1 and 2 and this will be equal to as when we solve as 1 minus 2 and 1 plus 2 and our value will be equal to 1 minus 2 is minus 1, 1 plus 2 is 3.

04:02 So here we have our first solution of the question answer of part a which is minus 1 and 3.

04:11 Now in the part b we will let here let z is equal to 2 times of 4.

04:21 X3 minus x4 plus 3 x5 which is equal to 2 minus 1 plus 3 plus 3 x5 that is equal to d of x to the 2 we can say that and therefore our vector z will be as vector 2x3 minus x 4 x5 plus 3 x5 will be equal to as v of d x of 2 is equal to d v x2 x2 into d d dash and that is as 2 minus 1 3 and here we have 12 2 minus 2 2 2 8 and 0 is 2, 0, n4, next multiplier is 2 minus 1 and 3.

05:41 So when we calculate this will be equal to as 16 minus 4 and 8.

05:52 And here we will have 2 minus 1 and 3.

05:56 So final calculation is 32 plus 4 plus 24 when we multiply these 2 together.

06:06 That is equal to 60...

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