What is the convolution of two independent standard gaussian distributions?

[stukbv]

Hello, my question ;

Suppose X1 and X2 are independent random variables, each with the standard gaussian distribution. Compute, using convolutions the density of the distribution X1 + X2 and show that X1+ X2 has the same distribution as X * root2 where X has standard gaussian distribution.

Basically I have said
take fx+y (z) = integral fx(z-y)fy(y) dy.

I have said fx (z-y) = 1/root2pi * exp(-.5(z-y)^2) and that fy = 1/root2pi exp(-(y^2)/2)

I have then multiplied these together inside an integral from minus infinity to infinity
I end up getting
1/2pi *exp(-z^2 / 2) * integral exp(y(-y+z))

Now how do i go further, and am i even on the right lines!?

Any help would be very much appreciated

Thanks.

 

[micromass]

Hi stukbv!

If I get it right, your problem is

\int_{-\infty}^{+\infty}{e^{-(y^2-zy)}dy}

Try to complete the square of y^2-zy...

 

[sabbagh80]

You are near the final result. consider micromass's hint.

if X,Y be independent RVs \widetilde{} G(0,1) then <br /> X+Y \widetilde{} G(0, \sqrt{2})

 

[stukbv]

Hi so now i have 1/2pi * e^(-k^2/2) * integral exp(-(l^2-kl))
Then i complete the square and get;
1/2pi exp(-k^2/4) integral exp(-(l-k/2)^2) dl

But how on Earth is this the same as the distribution of the other!?

Thanks

 

[micromass]

Well, you just need to calculate

\int_{-\infty}^{+\infty}{e^{-(y-z/2)^2}dy}

Isn't this easy to calculate? It is a well known integral... hint: substitute u=y-z/2

 

[sabbagh80]

Dear stukbv,
Don't forget: "integral of any pdf over all its range is equal to 1"!

 

[stukbv]

I still don't get it, i can integrate it normally but I wouldn't know how to work out the value at infinity and - infinity!?

 

[micromass]

What is

\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx}

Hint, it is the integral of a pdf...

 

[stukbv]

This is 1, but i don't have that above i have just exp(-u^2) with no / 2 ?!

 

[micromass]

Well, try a suitable substitution that will give you /2 in the exponent.

 

[stukbv]

:S how can i just change it ?

 

[micromass]

Just try a good substitution x=ay. What value must a be to get /2 in the exponent?

 

[stukbv]

x=2y-z?

 

[stukbv]

no, x= root2 (y-z/2)

 

[micromass]

no, x= root2 (y-z/2)

Yes! That looks good!

 

[stukbv]

So, i finally get that fx+y(z) = 1/root(2pi) *exp(-z^2/4)

And now to say that it equals dist root2 X where X ~ N(0,1) then do i just say that
D = root2 * X + 0 , so i can just replace all X's with D/root2 and its the same thing?
Does it matter that one uses D , i.e ill get the same as above but with D's where the Z's are, is this still ok?

 

[micromass]

So, i finally get that fx+y(z) = 1/root(2pi) *exp(-z^2/4)

And now to say that it equals dist root2 X where X ~ N(0,1) then do i just say that
D = root2 * X + 0 , so i can just replace all X's with D/root2 and its the same thing?
Does it matter that one uses D , i.e ill get the same as above but with D's where the Z's are, is this still ok?

This is quite difficult to follow. Can't you just say that X+Y\sim N(0,\sqrt{2})...

 

[stukbv]

i guess so yeah! thanks.

 

[stukbv]

Also , in another part it asks me to let Y=aX+b and says what is the density of the distribution of Y if X has the standard distribution, (i.e. to derive it) is it right to let Y-b/a = X, put these in place of all the x's in the fx and then multiply by the "stretch factor" of 1/a?