[Date Index] [Thread Index] [Author Index]
Re: TransformedDistribution, for a sum of M iid variables
On 7/21/11 at 9:07 PM, paulvonhippel at yahoo.com (paulvonhippel at yahoo) wrote: >Using the TransformedDistribution function it is easy to show that >the sum of two normal variables is normal, e.g., <code snipped> >Likewise it wouldn't be hard to show that the sum of 3 normal >variables is normal, or 4. >But how do I show that the sum of M normal variables is normal, >where M is an arbitrary positive integer. Rather than approach this problem using TransformedDistribution, I would approach this using MomentGeneratingFunction. The moment generating function for the sum of n things selected from a given distribution is the same as the moment generating function for the distribution raised to the nth power. That is for the sum of n standard normal deviates the moment generating function is: In:= MomentGeneratingFunction[NormalDistribution, t]^n Out= (E^(t^2/2))^n Comparing this to In:= MomentGeneratingFunction[NormalDistribution[0, s], t] Out= E^((s^2*t^2)/2) it is clear the distribution for the sum of n deviates drawn from a standard normal distribution is a normal distribution with mean = 0 variance = n. I should point out the moment generating function doesn't exist for all distributions. In those cases you can use the characteristic function which always exists. And since the characteristic function is essentially the Fourier transform of the distribution function, you can use the inverse Fourier transform to recover the distribution function from the characteristic function. Use of the characteristic function or moment generating function is a very useful technique for answering these types of questions. Each distribution has a unique characteristic function (and moment generating function when it exists). Also, you can easily compute the moments of the distribution directly from the characteristic function or moment generating function without actually finding the distribution function.