Covariance of two dependent variables

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jegues
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Homework Statement



See figure attached

Homework Equations





The Attempt at a Solution



I am not concerned with part (a), I have deduced that indeed X and Y are dependent.

I'm not sure if I have done part (b) correctly, and I am quite certain I have done part (c) incorrectly, but I couldn't think of what else to do.

Am I on the right track in parts (b) and (c)? What is incorrect, or how should I approach the problem?

Thanks again!
 
Attachments
  • Text of homework problem 2 on a full-wave rectifier: Y = |X| + N with Gaussian noise N, followed by parts (a) to (d) on dependence, covariance and correlation of X and Y.
    ProbQ2.JPG
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  • Handwritten solution on graph paper: a block diagram X to rectifier to Y = |X| + N with Gaussian noise pdf, a uniform pdf on [a, b], conditional pdf p(y | |X| = c) sketch, and computation of E[X] = 0 for X uniform on [-1, 1].
    Q-_0181.jpg
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  • Handwritten derivation on lined paper of covariance for dependent variables: E{Y} = E{|X|+N} = |X| = mu_Y, then sigma_XY = E{(X-mu_X)(Y-mu_Y)} simplified using E{X N} = 0, and part c with X ~ U(0,2) computing E{X} = 1 and sigma_XY = E{X(|X|+N)} - mu_X mu_Y.
    Q-_0182.jpg
    50.3 KB · Views: 625
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Here is the last page of my attempt.
 
Attachments
  • Handwritten derivation on lined paper of the covariance sigma_XY as E[X^2] + E[XN] - X, expanded with sigma_X^2 and mu_X^2 terms, then X ~ U(0,2) density is 1/2 on 0 to 2 and sigma_X^2 = 1/3, giving sigma_XY = 1/3 + 1 - X.
    Q-_0183.jpg
    43.5 KB · Views: 495
B) looks ok. you should expect zero covariance. The two are clearly anticorrelated where X is negative and, equally, correlated where X is positive.
In c), something strange at the bottom of the first sheet. How did μXμY become X?
 
haruspex said:
B) looks ok. you should expect zero covariance. The two are clearly anticorrelated where X is negative and, equally, correlated where X is positive.
In c), something strange at the bottom of the first sheet. How did μXμY become X?

I calculated μx for that case, it was found to be 1.

Since Y = |X| + N, and X is in the range of (0,2) |X| = X.

Thus, Y = X + N, I figured this pdf would simply be the pdf of N (it is known to be Gaussian) shifted to the right by X, having a mean value of X. Hence, μy = X.

I don't think that is correct, because I think μy should be a number.

Where did I go wrong? How do I fix part (c)?
 
jegues said:
I don't think that is correct, because I think μy should be a number.
Quite so. E[Y] = E[|X|+N] = E[|X|]+E[N].
 
haruspex said:
Quite so. E[Y] = E[|X|+N] = E[|X|]+E[N].

Okay so from this,

[tex]E[Y] = E[X] + 0 = \mu_{X} = 1 = \mu_{Y}[/tex]

but I am still stuck in finding,

[tex]\sigma_{XY} = E[(X-\mu_{X})(Y-\mu_{Y})] = E[(X-1)(Y-1)] = E[(X-1)(|X|+N-1)][/tex]

Any ideas?