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Convolution of two independent normally distributed random variables

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Wenjuan
Wenjuan 2012-9-12
评论： Torsten 2019-3-7
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);

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回答（3 个）

Wayne King
Wayne King 2012-9-14
编辑：Wayne King 2012-9-14
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
Rick Rosson
Rick Rosson 2012-9-14
Please try:
mean(x)
std(x)
What does it show? Is it what you expected? Why or why not?
Likewise:
mean(y)
std(y)
Star Strider
Star Strider 2012-9-14
编辑：Star Strider 2012-9-14
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);

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