Squaring Normal Distribution

**theV** · May 8th 2011, 05:07 AM

Hi all,

If I have one variable that is standard-normal distributed, say X ~ N(0,1) and another variable Y = X^2. What would the mean and variance of Y be?

I tried looking at a graph of x^2 vs. x^4 to try and think about this more intuitively. The result is that the x^4 graph is flatter for longer (i.e. lower gradient for more values of x) but I'm struggling to work out the implications this would have for a normal distribution.

Cheers.

**mr fantastic** · May 8th 2011, 05:20 AM

Originally Posted by theV

Hi all,

If I have one variable that is standard-normal distributed, say X ~ N(0,1) and another variable Y = X^2. What would the mean and variance of Y be?

I tried looking at a graph of x^2 vs. x^4 to try and think about this more intuitively. The result is that the x^4 graph is flatter for longer (i.e. lower gradient for more values of x) but I'm struggling to work out the implications this would have for a normal distribution.

Cheers.

You obviously need to calculate $E(X^n) = \int_{-\infty}^{+\infty} x^n f(x) \, dx$ where n = 2 and 4 and then use the usual definition of expected value and variance. And life is easy if you just get these expected values from the moment generating function of X.

Alternatively, the pdf of Y is well known ( see Chi-square distribution - Wikipedia, the free encyclopedia) and therefore so are E(Y) and Var(Y).

**theV** · May 8th 2011, 05:33 AM

Thanks for your reply. However, I don't quite understand how your response fits in with my query.

Where does the n=2 and n=4 come into it? That was just me thinking out loud with another example (i.e. the second paragraph in my original post has nothing to do with what I want to solve, I was just trying to show my thought process).

**SpringFan25** · May 8th 2011, 06:07 AM

the integral Mr F showed you is the definition for the mean of $X^k$ .

You can get the mean quickly with the following, which is true for any random variable:
$Var(X) = E(X^2) - \left(E(X)\right)^2$

Subsititute known values: Var(X)=1 and E(X) = 0
$1 = E(X^2) - 0$
$1 = E(Y)$
Which tells you the mean of Y is 1.

Variance of Y
$Var(Y) = E(Y^2) - \left(E(Y) \right)^2$
$Var(Y) = E(Y^2) - 1^2$
$Var(Y) = E(Y^2) - 1$
$Var(Y) = E((X^2)^2) - 1$
$Var(Y) = E((X^4)) - 1$

The last line should explain why Mr F suggests you might be interested in E(X^4).

**theV** · May 8th 2011, 08:39 AM

Thanks SpringFan - getting there slowly!

I understand all the working you posted. E[X] can be found using the integral - but why n=2 AND 4? Should it not only be n=4?

Any links to useful background reading would be appreciated.

**SpringFan25** · May 8th 2011, 08:47 AM

I didn't use n=2 because i'd already worked it out another way in the previous part of my post. It would be perfectly valid to write:

$Var(Y) = E(Y^2) - \left(E(Y) \right)^2$
$Var(Y) = E((X^2)^2) - (E(X^2))^2$
$Var(Y) = E(X^4) - (E(X^2))^2$

in which case you need n=4 and n=2.

As for background reading, i dont know. I use google and wikipedia, although some wiki stats articles contain mistakes so dont take them too seriously. Mr F gave you a link to the wiki article on the chi sqaure distribution. You may want to read that and see if you can convince yourself that $X^2$ follows a chi square distribution with parameter 1. (You dont need to know any deep maths to see that this is the case, just read the definition of the distribution carefully).

**theV** · May 8th 2011, 09:16 AM

Ok so in that case, my calculation for the Var(Y) won't be a definite integer, instead it will be:

Var(Y) = 1/5 x^5 - 1 and E(Y) = 1

Correct?

**SpringFan25** · May 8th 2011, 09:20 AM

i think you have tried to integrate $\int_{-\infty}^{\infty} x^4 f(x) dx$ and used the normal rule of integration for $x^4$ .

If thats the case, good try, but not correct. the f(x) makes the problem far more complicated than that. f(x) is a function of x called the pdf, for the normal distribution integral you need to do is:

$E(X^4) = \int_{-\infty}^{\infty} x^4 f(x) dx$
$E(X^4) = \int_{-\infty}^{\infty} x^4 \frac{1}{\sqrt{2\pi\sigma^2}}\; e^{ -\frac{(x-\mu)^2}{2\sigma^2} } dx$

Judging from the tone of your posts and the forum you are posting in, you're either at highschool or on an introductory stats course at college. Im not familiar with your education system, but i doubt you're expected to tackle that integral.

**theV** · May 8th 2011, 09:27 AM

How do I go about calculating E(X^4)?

**SpringFan25** · May 8th 2011, 09:32 AM

Evaluating the above integral is difficult and i wouldn't encourage you to try.

Mr F already suggested a method for finding E(X^4), which was to use the moment generating function. However if you dont recognise that term, it's probably beyond your level of study.

Can you convince yourself that Y follows a chi square distribution with parameter 1? if so, just look up the variance of the chi square distribution. (this was also suggested by Mr F in his first post)

**theV** · May 8th 2011, 11:19 AM

Your last two posts have been fantastic, SpringFan!

My new dilemma is this: I can't convince myself that Y follows a chi square distribution with just 1 degree of freedom.

For others in a similar position to me, here is a good intro video to the Chi square distribution:

**mr fantastic** · May 8th 2011, 12:17 PM

Originally Posted by theV

Your last two posts have been fantastic, SpringFan!

My new dilemma is this: I can't convince myself that Y follows a chi square distribution with just 1 degree of freedom.
[snip]

Then you're not expected to know that either because you don't have the mathematical tools to do the calculation. Use Google to find a proof.

It would help if you said why you asked the original question and gave your mathematical background. Then we could (most likely) say that you don't yet have the mathematical knowledge to answer it and a lot of time would be saved.

**theV** · May 8th 2011, 12:24 PM

Originally Posted by mr fantastic

It would help if you said why you asked the original question and gave your mathematical background. Then we could (most likely) say that you don't yet have the mathematical knowledge to answer it and a lot of time would be saved.

Sure - this is for an intro econometrics module at university. I'm fairly certain the chi-square approximation is the correct way of doing it, but all the sources I have found say that normal ~ chi-square only for larger degrees of freedom - not df = 1. What else can I consider?

**mr fantastic** · May 8th 2011, 12:29 PM

Originally Posted by theV

Sure - this is for an intro econometrics module at university. I'm fairly certain the chi-square approximation is the correct way of doing it, but all the sources I have found say that normal ~ chi-square only for larger degrees of freedom - not df = 1. What else can I consider?

I gave you a link!! The distribution is exact. You have been told how to do the question. Clearly you are lacking the mathematical knowledge to prove what you've been told (and I doubt an eco intro module would require proofs anyway), that does not stop you from quoting the necessary results and using them.

I see no point in further questions until you take on board what has already been said.

**SpringFan25** · May 8th 2011, 12:32 PM

yay! another economist!

for the purposes of introductory econometrics, you can (probably) treat the following as the definition of a chi square distribution:

"take K independent standard normal variables, and square each of them. Add the results up. The sum follows a chi squared distribution with parameter k".

let k=1 in the above definition:
"take 1 standard normal variables, and square it. This follows a chi squared distribution with parameter 1".

You started by defining X as a standard normal variable, so it fits exactly into that definition.

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