I Correlation coeff in conditional distribution

[georg gill]

Can someone derive: $\frac{Cov(Z+\Theta),\Theta)}{\sqrt{Var(Z+\Theta)Var(\Theta)}}=\frac{\sigma ^2}{\sqrt{1+\sigma ^2}}$

My attempt:

Numerator:

$Cov(X,Y)=E[(X-E(X))(Y-E(Y))]=E[(Z+\Theta-\theta)(\Theta-\mu)]$

The denumerator is pretty simple:

$\sqrt{(1+\sigma ^2)\sigma ^2}$

 

[StoneTemplePython]

I presume you actually want $\rho$ which has a value of $\frac{\sigma}{\sqrt{1 + \sigma^2}}$ not $\frac{\sigma^2}{\sqrt{1 + \sigma^2}}$

key ideas:
1.) Break this up into small manageable subproblems

2.) Remember implications of independence between $Z$ and $\Theta$, i.e. they have zero covariance which also means their combined variance is just $var(Z) =1$ plus $var(\Theta) = \sigma^2$

-- numerator --
in general for covariance of two random variables, A,B, we have
$cov\big(A, B\big) = E[(A)(B)] - E[(A)]E[(B)]$

split this up into two lines

$E[(Z + \Theta)(\Theta)] = E[Z \Theta] + E[\Theta^2]$
$E[Z + \Theta]E[\Theta] = E[Z]E[\Theta]+ E[\Theta]E[\Theta]$

notice that our actual expression for the numerator is the first line minus the second

$= \big(E[Z \Theta] + E[\Theta^2]\big)- \big(E[Z]E[\Theta]+ E[\Theta]E[\Theta]\big) =\big(E[Z \Theta]- E[Z]E[\Theta]\big) + \big(E[\Theta^2]- E[\Theta]E[\Theta]\big)$
$= \big(cov(Z,\Theta)\big) + \big(var(\Theta)\big) = \big(0\big) + \big(\sigma^2\big) = \sigma^2$

-- denominator --

$\sqrt{\Big(\big(1 + \sigma^2\big)\big(\sigma^2\big)\Big)} = \sqrt{\big(1 + \sigma^2\big)}\sqrt{\big(\sigma^2\big)} = \sigma\sqrt{\big(1 + \sigma^2\big)}$--- combine numerator and denominator --

$\rho = \frac{\sigma^2}{\sigma\sqrt{\big(1 + \sigma^2\big)}} = \frac{\sigma}{\sqrt{\big(1 + \sigma^2\big)}}$

 

[georg gill]

I presume you actually want $\rho$ which has a value of $\frac{\sigma}{\sqrt{1 + \sigma^2}}$ not $\frac{\sigma^2}{\sqrt{1 + \sigma^2}}$

##

Yes sorry I meant $\rho$ which has a value of $\frac{\sigma}{\sqrt{1 + \sigma^2}}$. Thanks for the answer!

 

[georg gill]

How do they arrive at the formula for expectation and variance in the end. I am thinking about:

$E[\Theta|X=x]=E[\Theta]+\rho\sqrt{\frac{Var(\Theta)}{Var(X)}}(x-E[X])$

and

$Var(\Theta|X=x)=Var(\Theta)(1-\rho^2)$

Where do they take this formulas from. Can someone derive them? I get the calculations they do with them.

 

[StoneTemplePython]

You should start by stating the relationship between $\Theta$ and $X$. I didn't think this was directly needed when dealing with $Z$ and $\Theta$ but it seems vital now and I don't see this clearly stated anywhere, and $\Theta$ and $X$ seem to be being introduced in your exercise 8b as if you have familiarity with the relationship from the text or perhaps example 8a. Consider this a 'for avoidance of doubt', the relationship is ____, type of statement.

After stating the relationship, it may be prudent to try to draw this out via a picture. Then make an attempt at solving this, either using their stated approach or via direct application of Bayes.

The reality is, once you've stated and formulated everything, the expected values should be easy -- and if you get stuck, following the units should be helpful here as well.

I could weigh in after all of this if needed. But you should be able to (a) clearly state the relationship between $\Theta$ and $X$ and (b) make some progress on your own first.

- - - -
edit:

While I still think a lot more details should be provided, I knew this equation looked quite familiar. Your problem apparently is trying to minimize Linear Least Mean Squared error. I.e. you are trying to minimize

$E\big[(\Theta - \alpha X - b)^2\big]$

do the calculus on this minimization problem (i.e. optimize with respect to a and b) and you'll recover the abstract form of your conditional expected value