# Math Help - Random Variable

1. ## 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

2. 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

3. does it make a difference that your paremeters are unknown while solving and also, the result of subscript 1 and n?

4. 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.

5. 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)

6. 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, ....