Convolution of two independent normally distributed random variables

If x and y are independent and both normal with mean=5, and v=4, then z=x+y should be normal with mean=10, and v=8. Why can't I proof this using convolution in Matlab.

a=linspace(-5,15,10000); x=normpdf(a,5,2); y=normpdf(a,5,2); z=conv(x,y); m=mean(z); sd=std(z);

Answer 1

[Wayne King]

Hi, you are confusing things here a bit. You don't want to consider the mean and standard deviations of the PDFs. You want to consider the mean and standard deviation of the random variables. Those are very different things.

x = normrnd(5,2,1000,1);
y = normrnd(5,2,1000,1);
z = x+y;
mean(z), var(z)

To see that the N(0,8) (here I mean variance 8) is good fit to z

 x = normrnd(5,2,1000,1);
 y = normrnd(5,2,1000,1);
 z = x+y;
 mu = 10;
 sigma = sqrt(8);
 zmin = min(z); zmax = max(z); zrange = range(z);
 binw = zrange / 30;
 edges = zmin + binw*(0:30);
 n = histc(z, edges);
 bar(edges, n, 'histc');
 hold on
 xx = zmin:(zrange/1000):zmax;
 plot(xx, binw*length(z)*normpdf(xx,mu,sigma), 'r');
 title('Normal Fit');

The PDF of z is the convolution of the PDFs of x and y

[Deepak Kumar]

what we will do if the z = x*y

[Torsten]

Does this help ?

http://mathworld.wolfram.com/NormalProductDistribution.html

Answer 2

[Rick Rosson]

Please try:

   mean(x)
   std(x)

What does it show? Is it what you expected? Why or why not?

Likewise:

   mean(y)
   std(y)

Answer 3

[Star Strider]

Since the convolution is based on the variables described by a specific distribution, it's likely best to look at the inverse normal distribution:

p = linspace(0.001,0.999,10000); 
x = norminv(p,5,2);
y = norminv(p,5,2);

These give the values of a mean of 5 and a standard deviation of 2:

mx = mean(x);
sx = std(x);
my = mean(y);
sy = std(y);

and these give the values of a mean of 10 and standard deviation of 4:

z = x + y;
mz = mean(z); 
sz = std(z);