Sum of two Standard Normal Random Variables.

[iocal]

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?

 

[DrPhil]

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?

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.

 

[iocal]

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.

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.