# Problem with calculating the cov matrix of X,Y

1. Jun 10, 2015

### ChrisVer

If I have two random variables $X, Y$ that are given from the following formula:
$X= \mu_x \big(1 + G_1(0, \sigma_1) + G_2(0, \sigma_2) \big)$
$Y= \mu_y \big(1 + G_3(0, \sigma_1) + G_2(0, \sigma_2) \big)$

Where $G(\mu, \sigma)$ are gaussians with mean $\mu=0$ here and std some number.

How can I find the covariance matrix of those two?

I guess the variance will be given by:
$Var(X) = \mu_x^2 (\sigma_1^2+ \sigma_2^2)$ and similarly for Y. But I don't know how I can work to find the covariance?
Could I define some other variable as :$Z=X+Y$ and find the covariance from $Var(Z)= Var(X)+Var(Y) +2 Cov(X,Y)$ ?
while $Z$ will be given by $Z= (\mu_x+\mu_y) (1+ G_1 + G_2)$?

Then $Var(Z)= (\mu_x+ \mu_y)^2 (\sigma_1^2+ \sigma_2^2)$

And $Cov(X,Y) = \dfrac{(\mu_x+\mu_y)^2(\sigma_1^2+ \sigma_2^2)- \mu_x^2 (\sigma_1^2+ \sigma_2^2) - \mu_y^2(\sigma_1^2+ \sigma_2^2) }{2}=\mu_x \mu_y(\sigma_1^2+ \sigma_2^2)$

Is my logic correct? I'm not sure about the Z and whether it's given by that formula.

Last edited: Jun 10, 2015
2. Jun 10, 2015

### Orodruin

Staff Emeritus
Why not simply apply the definition of covariance? Or the reduced form E(XY)-E(X)E(Y)?

3. Jun 11, 2015

### ChrisVer

I don't know E[XY]...?

4. Jun 11, 2015

### Orodruin

Staff Emeritus
But you know the distributions of $X$ and $Y$ and so you can compute it.

5. Jun 11, 2015

### ChrisVer

I don't know the joint distribution function...
so that $E[xy]= \int dxdy~ h(x,y) xy$
And I can't say $h(x,y)=f(x)g(y)$ since I don't know if X,Y are independent...if they were independent they wouldn't have a covariance too...the E[xy]=E[x]E[y] and your reduced form formula would vanish the cov.

6. Jun 11, 2015

### Orodruin

Staff Emeritus
You do. The only reason I can see to number the $G$s is to underline out that they are independent distributions so that $X$ and $Y$ have a common part $G_2$ and one individual part $G_1$/$G_3$.

7. Jun 11, 2015

### ChrisVer

Yes that's the reason of labeling them... It just happens that $G_1,G_2$ have the same arguments $\mu=0, \sigma_1$ but they are not common for X,Y. However the $G_3$ is a common source of uncertainty in both X,Y... Still I don't understand how to get the joint probability from it... like taking the intersection of X,Y is leading to only the $G_3$? that doesn't seem right either.

8. Jun 11, 2015

### Orodruin

Staff Emeritus
I suggest you start from the three-dimensional distribution for $G_i$. From there you can integrate over regions of constant $X$ and $Y$ to obtain the joint pdf for $X$ and $Y$. In reality, it does not even have to be that hard. Just consider $X$ and $Y$ as functions on the three dimensional outcome space the $G_i$ and use what you know about those, e.g., $E(G_1G_2) = 0$ etc.

9. Jun 11, 2015

### Stephen Tashi

To keep the discussion clear, you should use correct notation. If "$X$" is a random variable then $X$ has a distribution, but it is not "equal" to its distribution. I think what you mean is that:

$X = \mu_x( X_1 + X_2)$
$Y = \mu_y( X_3 + X_2 )$

where $X_i$ is a random variable with Gaussian distribution $G(0,\sigma_i)$.

10. Jun 11, 2015

### ChrisVer

No, X is a random variable,taken from a distribution function with:
mean $\mu_x$ (that's the role of 1)
some measurment uncertainty following a guassian $G_1$ or $G_3$.
a further common measurement uncertainty $G_3$.

That means I could take 5 measurements from X :$\{x_1,x_2,x_3,x_4,x_5 \}=\mu_x+ \{ + 0.02, + 0.001, 0, - 0.01, - 0.06\}$.

11. Jun 11, 2015

### ChrisVer

Ahhh OK I understood what you meant to say... yes it's fine, I understand what I wrote maybe I didn't write it in a correct way (confusing random variables with distributions)...

12. Jun 11, 2015

### Stephen Tashi

Apply the formula for $\sigma(aX + bY, cW + dV)$ given in the Wikipedia article on Covariance http://en.wikipedia.org/wiki/Covariance

In your case , $a = b = \mu_x,\ X=X_1,\ Y= X_2,\ c=d=\mu_y,\ W = X_3, V = X_2$

Edit: You'll have to put the constant 1 in somewhere. You could use X = 1 + X_1.

13. Jun 11, 2015

### ChrisVer

writing the covariance as $\sigma_{XY} = \sigma_X \sigma_Y \rho$ with $\rho$ the correlation coefficient.

And since $X= \mu_x (1+ X_1 + X_2)$ and $Y = \mu_y (1 + X_3 + X_2)$ I think these two are linearly correlated (due to $X_2$). So $\rho>0$. Would you find this a logical statement? I mean if $X_2$ happens to be chosen a larger number, both $X,Y$ will get a larger number as contribution.
For the value of $\rho$ I guess it should (by the same logic) be given by a combination of $\mu_y,\mu_x$, since they give the difference in how $X,Y$ change with $X_2$'s change. I mean if $\mu_x > \mu_y$ then $X$ will get a larger contribution from the same $X_2$ than $Y$ and vice versa for $\mu_x<\mu_y$...So I guess it should be an $X(X_2)=\frac{\mu_x}{\mu_y} Y(X_2)$?

Last edited: Jun 11, 2015
14. Jun 11, 2015

### Orodruin

Staff Emeritus
I still think you are overcomplicating things. Do you know how to compute $E(X_iX_j)$ when $X_i$ has a Gaussian distribution with mean zero?

15. Jun 11, 2015

### ChrisVer

In general I'd know, but as I said I have difficulty in finding the joint pdf...

Are you saying that $\mu_i (1+G_2)$ in my notation is the joint pdf (after integrating out G1 and G3)?

16. Jun 11, 2015

### Orodruin

Staff Emeritus
You do not need to find the joint pdf. You can work directly with the Gaussians!

17. Jun 11, 2015

### ChrisVer

Then in that case I don't know how to find $E[X_i X_j]$....
The formula I know defines the expectation value through an integral with the pdf...

18. Jun 11, 2015

### Orodruin

Staff Emeritus
That one is simple, it is the expectation value of two independent gaussians with zero mean. Since they are independent, the pdf factorises ...

Edit: ... unless, of course, if i = j ...

19. Jun 11, 2015

### ChrisVer

So you suggest something like:
$E[X_i X_j] = \begin{cases} E[X_i]E[X_j] = \mu_i \mu_j & i \ne j \\ E[X_i^2]= \sigma_i^2 + \mu_i^2 & i=j \end{cases}$

Where $X_i$ gaussian distributed variables

20. Jun 11, 2015

### Orodruin

Staff Emeritus
Indeed.

Edit: Of course, you now have $E[(1+X_1+X_2)(1+X_3+X_2)]$ so you will have to do some algebra, but not a big deal.