# Thread: Squareing a Normal[0,1] random variable

1. ## Squareing a Normal[0,1] random variable

A certain random variable X has a a normal distribution with \[Mu] = 0 and \[Sigma] = 1. Come up with a formulas for the cumulative distribution function and probability density function of X^2 and plot each separately.

Would this be right ? Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]].

2. Originally Posted by aeubz
A certain random variable X has a a normal distribution with \[Mu] = 0 and \[Sigma] = 1. Come up with a formulas for the cumulative distribution function and probability density function of X^2 and plot each separately.

Would this be right ? Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]].

3. I'm not understanding it very much.. I need to know if this is the probability: Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]].

4. That's correct, but there is no closed form for the cdf of N(0,1)
You have to use the transformation of the pdf.

5. Originally Posted by Moo
That's correct, but there is no closed form for the cdf of N(0,1)
You have to use the transformation of the pdf.
He can do what he is doing and just take the derivative of the expression he already has to get the pdf. Transformation method is fine too, but some intro texts don't cover transformations that are only monotone on intervals (mine didn't, at least).

If $Y = X^2$, you've already derived $F_Y(y) = F_X (\sqrt{y}) - F_X (-\sqrt{y})$. Just take the derivative with respect to y to get the pdf of Y in terms of the pdf of X, which you have a closed form expression for. You should recognize the form of the pdf straight away and know that there isn't a closed form formula for the cdf.

6. Originally Posted by aeubz
I'm not understanding it very much.. I need to know if this is the probability: Prob[X^2 <= x] = Prob[-Sqrt[x] <= X <= Sqrt[x]] = Xcdf[Sqrt[x]]- Xcdf[-Sqrt[x]].