# Homework Help: Show that X+Y is normal

1. Aug 28, 2010

### Gekko

1. The problem statement, all variables and given/known data

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 cant find it anywhere. Simply adding the marginal distributions, X+Y is what I tried but how is the correlation coefficient introduced?

2. Aug 28, 2010

### lanedance

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

3. Aug 28, 2010

### 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?

4. Aug 29, 2010

### Gekko

Unfortunately this approach doesnt seem to work. I end up with a very messy exponential which doesnt allow separation for the marginal calculation :(

Any thoughts? Is this not a standard proof?

5. Aug 29, 2010

### 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

6. Aug 29, 2010

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.

7. Aug 30, 2010

### 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

8. Aug 30, 2010

### 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

9. Aug 30, 2010

### lanedance

note if they have mean 0 & variance 1 the covariance matrix is
$$\Sigma = \begin{pmatrix} 1 & \rho \\ \rho & 1 \end{pmatrix}$$

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

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

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

Last edited: Aug 30, 2010
10. Aug 30, 2010

### lanedance

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

Last edited: Aug 30, 2010
11. Aug 30, 2010

### 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 doesnt 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...

12. Aug 30, 2010

### 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

13. Aug 30, 2010

### 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?

14. Aug 30, 2010

### Gekko

15. Aug 30, 2010

### 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

16. Aug 30, 2010

### Gekko

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

17. Aug 30, 2010

no worries