# Thread: Moment Generating Function of a Chi-Squared distribution

1. ## Re: Moment Generating Function of a Chi-Squared distribution

Originally Posted by chippy
@Dragan : Hi Dragan, I would just like to know where the "obvious" part comes into play? A more direct route is always welcome.
If a random variable X has an exponential distribution with a mean and standard deviation of 2, then its probability density function is chi-square with df=2.

2. ## Re: Moment Generating Function of a Chi-Squared distribution

Once again, if I read your implication,
without normalizing them, the given distribution N(0,1) has...1 degree of freedom
, I'm not completely sure if we're on the same page. Your statement is incorrect, but I'm not 100% sure what you are actually saying, so I'll start from the beginning : That N(0,1) you are referring to - that is the distribution that results after standardizing (i'm very sorry - I've been using the word "normalize" in my previous posts) a normal random variable (which yields another normal distribution) and the 1 at the end is the variance of the distribution - not degrees of freedom.

To go down the traditional path of evaluating the probability in question, I would say "yes, as a starting point, you do need to normalize them". This was indeed my first reaction and by the looks of it, that of BGM as well. I'm interested though in hearing what Dragan's viewpoint in this is, seems there is a more direct route.

To use elementary techniques we needs to find distributions for and . We can use fact number (2) in BGM's post only if the conditions of number (2) are satisfied, which is that the RV must be normally distributed with mean 0 and variance of 1 (i.e. ) - therefore the standardization. The resulting chi-squared distribution has 1 degree of freedom - this, also, coming from fact (2) in BGM's post.

3. ## The Following User Says Thank You to chippy For This Useful Post:

saternal (05-09-2012)

4. ## Re: Moment Generating Function of a Chi-Squared distribution

If a random variable X has an exponential distribution with a mean and standard deviation of 2, then its probability density function is chi-square with df=2
Yes, I got that - not quite the part I thought you were referring to. I was wondering about the
theta = 8
part.

5. ## Re: Moment Generating Function of a Chi-Squared distribution

I was wondering about the...
theta = 8
part.

Well, 4 + 4 = 8.

6. ## Re: Moment Generating Function of a Chi-Squared distribution

Originally Posted by chippy
Once again, if I read your implication, , I'm not completely sure if we're on the same page. Your statement is incorrect, but I'm not 100% sure what you are actually saying, so I'll start from the beginning : That N(0,1) you are referring to - that is the distribution that results after standardizing (i'm very sorry - I've been using the word "normalize" in my previous posts) a normal random variable (which yields another normal distribution) and the 1 at the end is the variance of the distribution - not degrees of freedom.

To go down the traditional path of evaluating the probability in question, I would say "yes, as a starting point, you do need to normalize them". This was indeed my first reaction and by the looks of it, that of BGM as well. I'm interested though in hearing what Dragan's viewpoint in this is, seems there is a more direct route.

To use elementary techniques we needs to find distributions for and . We can use fact number (2) in BGM's post only if the conditions of number (2) are satisfied, which is that the RV must be normally distributed with mean 0 and variance of 1 (i.e. ) - therefore the standardization. The resulting chi-squared distribution has 1 degree of freedom - this, also, coming from fact (2) in BGM's post.
What is the meaning of N~[0,4] at the start then? Is it showing us the variance and mean of U , V ?

Sorry but i'm trying my best to fully understand the question and meaning of these concepts.

7. ## Re: Moment Generating Function of a Chi-Squared distribution

Originally Posted by saternal
What is the meaning of N~[0,4] at the start then? Is it showing us the variance and mean of U , V ?
Yes, you are correct. I think in this case you should first google "Normal Distribution" and read a bit about it, as well as standardizing the variable - just to understand the basics of this distribution first, before tackling the transformation in your original post.