# Random Variable

• Sep 26th 2008, 02:48 PM
weakmath
Random Variable
1. Suppose that X ~ N (mu, sig^2). Show

((x-mu)/sig)^2 ~ Gamma(1/2, 1/2.) = Ki-square subscripte 1

2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n
• Sep 26th 2008, 03:34 PM
mr fantastic
Quote:

Originally Posted by weakmath
1. Suppose that X ~ N (mu, sig^2). Show

((x-mu)/sig)^2 ~ Gamma(1/2, 1/2.) = Ki-square subscripte 1

2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n

This thread contains the ideas you need to use: http://www.mathhelpforum.com/math-he...tml#post119542
• Sep 26th 2008, 03:39 PM
weakmath
does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?
• Sep 26th 2008, 03:49 PM
mr fantastic
Quote:

Originally Posted by weakmath
does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?

Your parameters are not unknown. The mean is $\mu$ and the standard deviation is $\sigma$. All you have to do is generalise the solution given in the thread I referenced.

The subscripts 1, 2, .... n are just part of the symbol used for the random variable.
• Sep 26th 2008, 04:25 PM
mr fantastic
Quote:

Originally Posted by weakmath
[snip]
2. Suppose that X1,X2.... Xm are iid N(mu, sig^2). Show that

summation i = 1 to n ( (Xi - mu)/sig)^2 ~ Ki-square subscript n

I'd suggest using a moment-generating function approach for Q2.

I don't have time now but if you're stuck, post your work and I'll give some details later.

(You could also use this approach for Q1)
• Sep 27th 2008, 05:28 AM
mr fantastic
Quote:

Originally Posted by weakmath
does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?

In fact, at second glance Q1 is even easier than I thought. The thread I referenced actually gives a solution to Q1 since

$\frac{X - \mu}{\sigma} = Z$ ~ $N(0, \, 1)$.

Q2:

The moment generating function of $Z^2$ is:

$M_{Z^2}(t) = E\left[e^{tZ^2}\right] = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty}e^{tz^2} e^{-z^2/2} \, dz = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} e^{-z^2(1 - 2t)/2} \, dz$ $= \frac{1}{(1 - 2t)^{1/2}}$

(which by the way is recognised as the moment generating function for the chi-squared distribution with 1 degree of freedom).

Therefore the moment generating function of $Z_{1}^2 + Z_{2}^2 + \, .... \, + Z_{n}^2 = M_{Z_{1}^2}(t) \cdot M_{Z_{2}^2}(t) \cdot \, .... \cdot M_{Z_{n}^2}(t) = \frac{1}{(1 - 2t)^{n/2}}$

(since $Z_{1}^2, \, Z_{2}^2, \, .... \, Z_{n}^2$ are i.i.d. random variables).

This is recognised as the moment generating function for the chi-squared distribution with n degrees of freedom. And since when a moment generating function exists there is a unique distribution corresponding to that moment generating function, ....