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Random vector. random vector X with zero mean has a covariance matrix Zx with the following eigendecomposition Zx {416 0.5 (2) 0 } What is the variance of each of ...

Question

Random vector. random vector X with zero mean has a covariance matrix Zx with the following eigendecomposition Zx {416 0.5 (2) 0 } What is the variance of each of the entries of the random vector X1, Xz and X3? b. Is it possible to find a unit-norm vector U such that the inner product between X and & (i.e. the amplitude of the projection of X onto that direction) has variance greater than 1? Find three constants 01, 42 and 43, such that at least one of them is nonzero and P(a1X} + (2Xz + a3

Random vector. random vector X with zero mean has a covariance matrix Zx with the following eigendecomposition Zx {416 0.5 (2) 0 } What is the variance of each of the entries of the random vector X1, Xz and X3? b. Is it possible to find a unit-norm vector U such that the inner product between X and & (i.e. the amplitude of the projection of X onto that direction) has variance greater than 1? Find three constants 01, 42 and 43, such that at least one of them is nonzero and P(a1X} + (2Xz + a3X3 = 0) = 1. Justify your answer mathematically; and interpret it geometrically:



Answers

Given multivariate data $\mathbf{X}_{1}, \ldots, \mathbf{X}_{N}$ (in $\mathbb{R}^{p} )$ in meandeviation form, let $P$ be a $p \times p$ matrix, and define $\mathbf{Y}_{k}=P^{T} \mathbf{X}_{k}$ for $k=1, \ldots, N$
a. Show that $\mathbf{Y}_{1}, \ldots, \mathbf{Y}_{N}$ are in mean-deviation form. $[\text { Hint: }$ Let $\mathbf{w}$ be the vector in $\mathbb{R}^{N}$ with a 1 in each entry. Then $\left[\begin{array}{lll}{\mathbf{x}_{1}} & {\cdots} & {\mathbf{X}_{N}}\end{array}\right] \mathbf{w}=\mathbf{0}\left(\text { the zero vector in } \mathbb{R}^{p}\right) \cdot ]$
b. Show that if the covariance matrix of $\mathbf{X}_{1}, \ldots, \mathbf{X}_{N}$ is $S$ then the covariance matrix of $\mathbf{Y}_{1}, \ldots, \mathbf{Y}_{N}$ is $P^{T} S P .$

Yeah. This problem we were told that sigma squared for X is fine Insert required for why is three. We would like to find the variance of the random variable dizzy. Doesn't mean you have to acts plus four. Why minus three? The variance of the it was a variance of negative two X plus four Y minus three. Yeah. Which using our themes from the book is the same as two squared times the variance effect Most four square in terms of variants of why. And then that constant on the end does not change the variance. And so we know the variance of that Is five. So this is four times 5 16 times of variants of Y which we know to be three. So this is 20 plus 48 Which is 68. It's our variants of Z is 68.

Well here we are, given a random variable X. That has a discreet uniform distribution On the into jersey fixed between one and 3 and first for us to determine the mean and the variance. So the mean and the various formulas are given here. So the meanest P plus K over two, which is basically wants to to experience, is the minus a plus one squared minus 1/12. So three squared is nine, 9 -1 is eight. 8/12, yes .667. So we have our mean and variance.

For this exercise, we have a random variable X. That has a discreet uniform distribution on the integers one through three. So in other words, the probability of X taking on either of the three values 12 or three is one third for each. We are asked in the question to determine the mean and the variance of X. The mean or expected value of X. When it is a uniform discrete variable is A plus B, divided by two where a is the smallest integer, and B is the largest integer, And that gives us a mean of two. Now to solve the variants, we use this formula, so we have three minus one plus one, All squared -1 all over 12. And this comes out to you to over three. And so this random variable with a discreet uniform distribution has a mean of two and a variance of 2/3.

In this question we have to find the normal vectors of a given plane and the occupation of land is given to us. That is too in two x minus that. It was too three into explains why we are going to simplify the situation. So we get here two x minus two, that it was 23 x plus three Y. After subtracting to accent adding to see what we got here, I can say that X-plus three way and Less of 2:30. The cost to zero. So this is the equation of plane that was provided to us. Now we have to find the normal vector from here. Okay, so we can see that what happens here is first of all normal that this must have the components or coefficient I, J and K. So I can say that the coefficients of I J and K in normal vector is always equals to the coefficients of X. Hawaii and said of your regulation of lin. So due to this reason, we can see that no, the normal that should be given by you can see that coefficient of X in the equation is one year. So we have to write one times of I and coefficient of Y is three here. So three times of J. Okay. And coefficient of that is to hear. So two times of K. And finally I can say that I got the normal weapon earlier. That is I and plus three J and plus. Okay, okay. This is the answer of the given question here. Thank you.


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