Sum of two Standard Normal Random Variables.

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iocal

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Hi guys,

Could you please check whether what I've done is correct in the following? Let x and y be independent random variables, each with a distribution that is N(0,1). Then let z=x+y. Find the integral that represents the distribution function \(\displaystyle \displaystyle P(x+y \leq z)\). Then determine the probability density function of z.

Okay for the distribution function it is easy to see that the integral takes the form: \(\displaystyle \displaystyle G(z)=\int_{-\infty}^{\infty}\int_{-\infty}^{z-x} \frac{1}{2\pi} e^{-(x^2+y^2)/2} dydx\). Differentiating w.r.t to z and using Leibniz's formula we then get: \(\displaystyle \displaystyle g(z)=\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{-[x^2+(z-x)^2]/2}dx=\frac{1}{2\pi}e^{-z^2/4}\int_{-\infty}^{\infty}e^{-(x-z/2)^2}dx\), after completing the square. Now letting w=x-z/2 and assuming that the RV in the integral has variance equal to \(\displaystyle \ \frac{1}{2}\), the integral equals 1 and thus we are left with \(\displaystyle \displaystyle g(z)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt2}e^{-z^2/4}\), i.e. a normally distrubted RV with 0 mean and 2 variance(the sum of the variances of x and y). Essentially what we needed to show.

Everything alright here?
 
Last edited:
iocal said:
Hi guys,

Could you please check whether what I've done is correct in the following? Let x and y be independent random variables, each with a distribution that is N(0,1). Then let z=x+y. Find the integral that represents the distribution function \(\displaystyle \displaystyle P(x+y \leq z)\). Then determine the probability density function of z.

Okay for the distribution function it is easy to see that the integral takes the form: \(\displaystyle \displaystyle G(z)=\int_{-\infty}^{\infty}\int_{-\infty}^{z-x} \frac{1}{2\pi} e^{-(x^2+y^2)/2} dydx\). Differentiating w.r.t to z and using Leibniz's formula we then get: \(\displaystyle \displaystyle g(z)=\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{-[x^2+(z-x)^2]/2}dx=\frac{1}{2\pi}e^{-z^2/4}\int_{-\infty}^{\infty}e^{-(x-z/2)^2}dx\), after completing the square. Now letting w=x-z/2 and assuming that the RV in the integral has variance equal to \(\displaystyle \ \frac{1}{2}\), the integral equals 1 and thus we are left with \(\displaystyle \displaystyle g(z)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt2}e^{-z^2/4}\), i.e. a normally distrubted RV with 0 mean and 2 variance(the sum of the variances of x and y). Essentially what we needed to show.

Everything alright here?
Click to expand...
The only statement I object to is "assuming that the RV in the integral has variance equal to \(\displaystyle \ \frac{1}{2}\)." Such an "assumption" is not justified. Perhaps what you should say is, "comparing the integral to a normal distribution with ..." or "the integrand has the form of...".

I would also like to see intermediate steps in the differentiation with respect to z. [I am assuming you did it right, since you got the right answer.]

Sorry I didn't reply the first time you posted this - I had thought about it, but didn't [still haven't] take the time to do it myself in detail. But it feels right.
 
DrPhil said:
The only statement I object to is "assuming that the RV in the integral has variance equal to \(\displaystyle \ \frac{1}{2}\)." Such an "assumption" is not justified. Perhaps what you should say is, "comparing the integral to a normal distribution with ..." or "the integrand has the form of...".

I would also like to see intermediate steps in the differentiation with respect to z. [I am assuming you did it right, since you got the right answer.]

Sorry I didn't reply the first time you posted this - I had thought about it, but didn't [still haven't] take the time to do it myself in detail. But it feels right.
Click to expand...

You have been very helpful in everything I've posted so far so please no need to apologise!
I get your first point, I'll phrase it like that from now on. The differentiation has no intermediate steps though as this is a direct application of Leibniz's formula, for differentiation under the integral sign with respect to the upper bound of integration which is a function of z, i.e g(z), its derivative is 1, thus I have omitted the multiplication. Here http://en.wikipedia.org/wiki/Leibniz_integral_rule.

I would say that the same result can be arrived at using the moment generating function, but doing it the long way can be very insightful.

Anyway, thank you.
 
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