# Proving a normal distribution

1. Aug 22, 2010

### Gekko

Z=(-1/sqrt(n)) * sum from k=1 to n of [1+log(1-Fk)]

Fk is a cumulative distribution function which is continious and strictly increasing.

Show that as n->infinity, Z converges to a normal distribution with mean 0 and var 1

---------------------

From Taylor series, log(1-x) = -sum from 1 to infinity of (x^n)/n but dont see how this can help at the moment
Ive been looking for anything around the summation of c.d.fs but havent found anything so I think Im unaware of a few theorems which are essential to solving this. Any help appreciated. Been working on this for hours with no success.

2. Aug 22, 2010

### lanedance

can you exlpain what you mean by Fk? and how it depends on k?

I understand its a cumulative distribution function, continuous & strictly increasing...

however I may be missing something as I can't see where the randomness is coming into Z - otherwise I'd be thinking something long the lines of central limit theorem...

Last edited: Aug 22, 2010
3. Aug 22, 2010

### Gekko

Sorry, didnt make that clear. Xk is a sequence of independent random variables and Fk is its associated cumulative distribution function.

So Fk is a sequence of cumulative distribution functions taking on values form 0 to 1. Can we simply look at it like this? And hence assume it is equivalent to a sequence of independent random variables with values 0 to 1?

4. Aug 22, 2010

### Gekko

Some identities that are useful:

Expected value of X = integration from 0 to inf of [1-F(t)] dt where F(t) is the cdf
I need to obtain exp(-x^2/2) from the sum and log. Integrating x will obtain x^2/2 and taking the anti-log will obtain the exponential.

Not sure how all this fits together though

5. Aug 23, 2010

### Gekko

After looking at this more, it seems the moment generating function approach is the way to go.
By obtaining the mgf and finding the mean and variance to be 0 and 1, we prove it is a normal distribution.
Does anyone know if this is acceptable?

6. Aug 23, 2010

### lanedance

i still don't understand your explanation of Fk

is there only a single cdf F(x)?

so for every Xk has the same cdf Fi(x) = P(Xi<=x)?

if so, then the map between Xk & Fk is one to one monotonic and the random varable Fk, becomes a uniform random variable between 0 and 1

7. Aug 23, 2010

### Gekko

No, there is a cdf for each x.

Fk(xk) for k=1 to n
F1(x1), F2(x2) etc

Sorry, it's the lack of latex that makes it hard to show subscript

8. Aug 23, 2010

### Gekko

I looked at taking the mgf thinking if I can show the mean to be 0 and variance 1 from this approach and hence prove normality

g(t) = 1(1/sqrt(n) sum(from 1 to inf) exp(tk) [1+log(1-Fk(Xk))]

but this approach goes nowhere

9. Aug 23, 2010

### lanedance

even if they're different you should be able to show any Fk(x=Xk) represents a uniform random variable on [0,1], by definition of the cdf - have you tried using that property?

10. Aug 23, 2010

### Gekko

As n tends to infinity, the cumulative distribution function Fk(Xk) tends to F(X)

I think this is ok. I don't see how to show this tends to a normal distribution though.

I think I'm making this much harder than it is. Don't see anything close in any textbook or on the web!