Is X+Y Normally Distributed with a Bivariate Normal Distribution?

[Gekko]

Homework Statement

If X and Y have a bivariate normal distribution with correlation p, show that X +Y is normally distributed


This seems like a pretty standard proof but can't find it anywhere. Simply adding the marginal distributions, X+Y is what I tried but how is the correlation coefficient introduced?

 

[lanedance]

you need to start wth the joint pdf, which will contain info about the correlation

 

[Gekko]

If I change variables, U=X+Y and V=X-Y and take the jacobian, I can then take the marginal to find f_U.

However, how do you change the standard deviation and means when you perform change of variables?

 

[Gekko]

Unfortunately this approach doesn't seem to work. I end up with a very messy exponential which doesn't allow separation for the marginal calculation :(

Any thoughts? Is this not a standard proof?

 

[lanedance]

if X & Y have bivariate normal distribution (correlated), they can be wrtten in terms of 2 independent normal variables, say U & V, say:
X = aU + bV
Y = cU + dV

it should follow that Z = X+Y = (a+c)U + (b+d)V is a sum of 2 independent normal distributions & thus is normal

so if you know (or can show) the first part, you're there

 

[statdad]

If I change variables, U=X+Y and V=X-Y and take the jacobian, I can then take the marginal to find f_U.

However, how do you change the standard deviation and means when you perform change of variables?

You don't need to change them. Write out the joint distribution of X and Y, perform the transformation you describe, substitute, and then integrate out V. The integration will require you to complete the square in the exponent, and rewrite things as a single density.

 

[Gekko]

Thanks a lot for your replies. After making the substitution, there just isn't a way to minimize (completing the square or otherwise) because we have uv terms with different divisors. Is the approach rather to take u=x+y and v=something else? Is the choice of v the key?
Has anyone actually ever done this proof?

Please see Wolfram alpha link. No minimizing was possible
http://www.wolframalpha.com/input/?...u-v)-c)^2/d^2-2p((0.5(u+v)-a)(0.5(u-v)-c))/bd

 

[lanedance]

if you find the linear combinations that diagonalises the covariance matrix, those represent the independent variables, and you can do as i suggeseted in 5

 

[lanedance]

note if they have mean 0 & variance 1 the covariance matrix is
<br /> \Sigma = \begin{pmatrix} 1 &amp; \rho \\ \rho &amp; 1 \end{pmatrix}<br />

with \textbf{x}^T = (x,y)^T the joint pdf is proportional to
<br /> f_{X,Y}(x,y) = e^{ \frac{1}{2(1-\rho^2)} ( \textbf{x}^T \Sigma \textbf{x} ) }<br />

then eignevectors of sigma, say \textbf{u}^T = (u,v)^T are the independent RVs, that is their Joint pdf can be written as:
<br /> f_{U,V}(u,v) = e^{ (\frac{u}{\sigma_u})^2} e^{ (\frac{v}{\sigma_v})^2}<br />

this should be easy to show what you require, however statdads may save a step

 

[lanedance]

or how about trying what statdad suggested
<br /> f_Z(z) \approx \int \int dx dy \delta (z-x-y) e^{-x^2 -y^2 + 2 \rho x y }<br />
<br /> = \int dx e^{x^2 -(z-x)^2 + 2 \rho x (z-x) }<br />
<br /> = \int dx e^{-2(1+\rho)(x^2-zx + z^2) }<br />
<br /> = \int dx e^{-2(1+\rho)(x^2-zx + (\frac{z}{2})^2+(\frac{z}{2})^2) }<br />
<br /> = \int dx e^{-2(1+\rho)(x-\frac{z}{2})^2} e^{-2(1+\rho)(\frac{z}{2})^2 }<br />

 

[Gekko]

Lanedance, Thanks for your replies.

I believe what Statdad was referring to was the joint PDF of a bivariate distribution i.e. http://upload.wikimedia.org/math/b/1/0/b10ecc56f758b2f94a953e7e1bd2f1c2.png

In proving that X+Y is normal and where X and Y are dependent, not sure how you ignored the standard deviation in the dirac delta approach? In any case, when performing the final integration wrt x from -inf to inf doesn't yield the correct answer (probably my error somewhere)

I understand the approach Statdad outlined from the joint pdf but the problem is in completing the square to obtain the form that will allow integration giving the error function...

 

[lanedance]

sorry, i assumed mean zero variance one, and dropped any constants for simplicity, i though you could add them in later if the method works

anyway, taking off where we let off
= \int dx e^{-2(1+\rho)(x-\frac{z}{2})^2} e^{-2(1+\rho)(\frac{z}{2})^2 }

now you can move the last z term outside of the x integral
=e^{-2(1+\rho)(\frac{z}{2})^2 } \int dx e^{-2(1+\rho)(x-\frac{z}{2})^2

the integrand is then just the pdf of a normal variable, so should yield the same constant (1 if it were correctly normalised) regardless of z value, the z term just shifts the mean of the distrubtion

 

[Gekko]

Hi Lanedance

Integrating the final part wrt x gives:

sqrt(pi)/sqrt(2p+2) * exp(-2(1+p)(z/2)^2)

The sqrt(2p+2) is correct as the correlation component of the summation. Does the rest look OK?

 

[Gekko]

Actually, no the answer is here

http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

So something isn't working out in both the exponential (should be dividing by 2(1+p)) and the coefficient in front of the exponential (should be 1/sqrt(2pi))

 

[lanedance]

as i said I didn't indcude any constants (they're only for normalisation only, its the form that is important),

so if you want to get it exactly right you need to start with a correctly normalised distribution

 

[Gekko]

I see. Understood. Thanks a lot for taking the time to answer my questions. Appreciate it

 

[lanedance]

no worries