Linear transformation

[island-boy]

after a series of computations, I was able to get the following matrix equation from the given of a problem:

$$\[\left( \begin{array} {ccc} W_1 \\ W_2 \end{array} \right)\] = \[\left( \begin{array} {ccc} \frac{\sigma_{11}}{\sqrt{\sigma_{11}^2 + \sigma_{12}^2}} & \frac{\sigma_{12}}{\sqrt{\sigma_{11}^2 + \sigma_{12}^2}} \\ \frac{\sigma_{21}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}} & \frac{\sigma_{22}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}}\end{array} \right)\] \[\left( \begin{array} {ccc} Y_1 \\ Y_2 \end{array} \right)\]$$

where Y1 and Y2 are independent processes.

the correlation of W1 and W2 was given as follows:

$$\rho = \frac{\sigma_{11}\sigma_{21} + \sigma_{12}\sigma_{22}}{\sqrt{\sigma_{11}^2 + \sigma_{12}^2}\sqrt{\sigma_{21}^2 + \sigma_{22}^2}}$$

Now what the problem asks is that I be able to show the following matrix equation to be true:

$$\[\left( \begin{array} {ccc} W_1 \\ W_2 \end{array} \right)\] = \[\left( \begin{array} {ccc} 1 & 0 \\ \rho & \sqrt{1 - \rho^2} \end{array} \right)\] \[\left( \begin{array} {ccc} Z_1 \\ Z_2 \end{array} \right)\]$$

where Z1 and Z2 are independent
and rho is the correlation of W1 and W2.

My question is:
can I just let
$$\sigma_{11} = 1$$
$$\sigma_{12} = 0$$

and just let
Y1 = Z1
Y2 = Z2?

cause if I did so, then the matrix equation that I want to prove is satisfied.
that is, the following are now true:
$$\rho = \frac{\sigma_{21}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}}$$
$$\sqrt{1 - \rho^2} = \frac{\sigma_{22}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}}$$

Is this a valid way of proving?

or should I have to find a matrix linear transformation to transfrom the first equation that I got into the required equation? cause if that's the way that it shouldbe done, then I'm not sure how to proceed about it.

thanks for the help.

 

[AKG]

after a series of computations, I was able to get the following matrix equation from the given of a problem:

$$\[\left( \begin{array} {ccc} W_1 \\ W_2 \end{array} \right)\] = \[\left( \begin{array} {ccc} \frac{\sigma_{11}}{\sqrt{\sigma_{11}^2 + \sigma_{12}^2}} & \frac{\sigma_{12}}{\sqrt{\sigma_{11}^2 + \sigma_{12}^2}} \\ \frac{\sigma_{21}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}} & \frac{\sigma_{22}}{\sqrt{\sigma_{21}^2 + \sigma_{22}^2}}\end{array} \right)\] \[\left( \begin{array} {ccc} Y_1 \\ Y_2 \end{array} \right)\]$$

Let W denote the left side of this equation, let S denote the 2x2 matrix on the right side, and let Y denote what you think it denotes.

$$\[\left( \begin{array} {ccc} W_1 \\ W_2 \end{array} \right)\] = \[\left( \begin{array} {ccc} 1 & 0 \\ \rho & \sqrt{1 - \rho^2} \end{array} \right)\] \[\left( \begin{array} {ccc} Z_1 \\ Z_2 \end{array} \right)\]$$

where Z1 and Z2 are independent
and rho is the correlation of W1 and W2.

Let R denote the 2x2 matrix on the right side of this equation, and let Z denote what you think it does.

So you have W = SY and you want to show that if W = RZ, then the entries of Z are independent. You can't go setting $\sigma _{ij}$ and Z_i to whatever you like. What kind of question would that be? Anyways, if you want to get W = RZ, then it's equivalent to get SY = RZ. In order to get this, what must Z be? Clearly, Z = R^-1SY. So look at R^-1SY, and show it's entires to be independent. Use the fact that the entries of Y are independent, and that the entries of W have correlation $\rho$.

 

[island-boy]

hi thanks for replying

and you want to show that if W = RZ, then the entries of Z are independent.

actually, what I need to show is that I can write W in the form RZ where Z are independent.
that is, defining W, S, Y, R, and Z below, and given W = SY where Y are independent, I must show that I can write W as W = RZ where Z are independent.

does what you have written still apply?
thanks again.

 

[AKG]

Yes. ,

 

[island-boy]

hi AKG,

after thinking it throrughly, I think I understand what you meant.
I only need to solve for Z = R^{-1}SY
and then I show the elements of Z are independent of each other (possibly by showing that their covariance is 0)

Thanks for you help! cheers!