# Statistics: sampling distributions related to normal distribution

• Jan 13th 2009, 08:36 PM
kingwinner
Statistics: sampling distributions related to normal distribution
http://www.geocities.com/asdfasdf23135/stat1.JPG
iid=independent and identically distributed

This is the model answer to a question which is unlikely to be wrong. I don't understand the part circled in red. How is that step justified? ~ is not the same as a equal sign, so can we still do that?
In other words, my question is "If c is a constant and cW follows the t-distribution with parameter 12, i.e. cW~t(12), does this imply that W~1/c t(12)?"
If so, WHY?
I hope that someone can help me out. Thanks a lot!
• Jan 14th 2009, 04:21 AM
mr fantastic
Quote:

Originally Posted by kingwinner
http://www.geocities.com/asdfasdf23135/stat1.JPG
iid=independent and identically distributed

This is the model answer to a question which is unlikely to be wrong. I don't understand the part circled in red. How is that step justified? ~ is not the same as a equal sign, so can we still do that?
In other words, my question is "If c is a constant and cW follows the t-distribution with parameter 12, i.e. cW~t(12), does this imply that W~1/c t(12)?"
If so, WHY?
I hope that someone can help me out. Thanks a lot!

Let $X$ have a pdf $f(x)$ and consider finding the pdf of $U = \alpha X$:

The cdf of $U$ is $G(u) = \Pr(U < u) = \Pr(\alpha X < u) = \Pr\left( X < \frac{u}{\alpha}\right) = \int_{-\infty}^{u/\alpha} f(x) \, dx$.

Therefore the pdf of $U$ is $g(u) = \frac{dG}{du} = \frac{d}{du} \int_{-\infty}^{u/\alpha} f(x) \, dx = \frac{1}{\alpha} f\left( \frac{u}{\alpha}\right)$.
• Jan 14th 2009, 06:24 AM
kingwinner
Quote:

Originally Posted by mr fantastic
Let $X$ have a pdf $f(x)$ and consider finding the pdf of $U = \alpha X$:

The cdf of $U$ is $G(u) = \Pr(U < u) = \Pr(\alpha X < u) = \Pr\left( X < \frac{u}{\alpha}\right) = \int_{-\infty}^{u/\alpha} f(x) \, dx$.

Therefore the pdf of $U$ is $g(u) = \frac{dG}{du} = \frac{d}{du} \int_{-\infty}^{u/\alpha} f(x) \, dx = \frac{1}{\alpha} f\left( \frac{u}{\alpha}\right)$.

Firstly, here we are assuming alpha>0, right?

Secondly, the f(u/alpha) seems to make things complicated. I still can't see why "If c is a constant and cW follows the t-distribution with parameter 12, i.e. cW~t(12), then W~1/c t(12)". Could you explain a little bit further?

Thank you!
• Jan 14th 2009, 11:56 AM
mr fantastic
Quote:

Originally Posted by kingwinner
Firstly, here we are assuming alpha>0, right?

Secondly, the f(u/alpha) seems to make things complicated. I still can't see why "If c is a constant and cW follows the t-distribution with parameter 12, i.e. cW~t(12), then W~1/c t(12)". Could you explain a little bit further?

Thank you!

On behalf of the mathematics (which is the cause of the trouble), I apologise that "the f(u/alpha) seems to make things complicated".

My argument needs to be changed if $\alpha < 0$ because of the necessary reversal in the inequality as a result of dividing by $\alpha < 0$. I should have stated that, I just assumed $\alpha > 0$ because that was your specific circumstance. I will let you check what the end-result will be.

I have given you a formula with a clear derivation of where it has come from. Apply the formula for your particular situation. In fact, feel free to use the derivation as a guide for deriving the formula for your own specific case.
• Jan 14th 2009, 08:08 PM
kingwinner
But I haven't learnt about the density function of the t-distribution in my class. Do we need it in this case?
• Jan 15th 2009, 11:06 PM
kingwinner
Now I am seeing another trouble...
If X is a random variable, then I know what 1/c X means, but t(12) is a distribution, what is the meaning of saying some random variable follows the distribution 1/c t(12)?
• Jan 16th 2009, 04:15 AM
mr fantastic
Quote:

Originally Posted by kingwinner
Now I am seeing another trouble...
If X is a random variable, then I know what 1/c X means, but t(12) is a distribution, what is the maning of saying some random variable follows the distribution 1/c t(12)?

I have given you a formula. It tells you what the pdf of $\alpha X$ is (for $\alpha > 0$) in terms of the pdf of X.

In your case the pdf of X is the pdf for the t-distribution. The meaning of $\frac{1}{\alpha} f\left( \frac{u}{\alpha}\right)$ is that it's the pdf of $\alpha X$. I really don't know what else to say.