[Math] Finding the PDF using the CDF Method

probabilityprobability distributionsprobability theoryrandom variablesstatistics

I am currently working through a series of lecture notes on functions of random variables and I have been given the following as an exercise.

Find the PDF of $X=Z^2$ given $Z$~$N(0,1)$ using the CDF method.

I was under the impression that the normal distribution was not easily integrable, which would make this question a bad candidate for the cdf method.

Is this thinking correct or am I missing something ?

Best Answer

Yes, your thinking is wrong since we don't actually need to evaluate the CDF. We can simply notice that $\Phi'(z) = F_Z'(z) = f_Z(z) = \phi(z)$, where $\Phi$ and $\phi$ are the standard normal CDF and PDF respectively.

Proceed directly, $$P(X\leq x) = P(Z^2\leq x) = P(|Z|\leq \sqrt{x}).$$

What can you say about $|Z|\leq \sqrt x$? Can you rewrite $P(|Z|\leq \sqrt x)$ in terms of $\Phi$?

Related Solutions

[Math] Plot the cdf and simulate a random variable (rv) with this cdf using the inversion method.

The issue is that the normalizing constant has been incorrectly specified in the exercise. It says 2 when it should in fact be 3. See below.

$$ \int_1^2c(x-1)^2dx=c\left[\frac{(x-1)^3}{3}\right]_1^2=\frac{c}{3}=1\Leftrightarrow c=3 $$

If you change this and run the following code:

X.sim <- function(S)1+(S)^(1/3)
S <- runif(100000)   
x <- X.sim(S)
max(x)

you will see that you get very close to 2, but you won't exceed it. Hope this helps!

[Math] Finding the CDF of a piecewise PDF

In general an indefinite integral should have a "$+C$" term. You actually want the cumulative distribution function $F(x)$ - the probability of seeing the value $x$ or below - to be the definite integral of the density function from $-\infty$ to $x$.

You want $F(-\infty)=0$ and $F(\infty)=1$ in general.

Since in your example the density is zero below $x=-1$ and above $x=1$, you want $F(-1)=0$ and $F(1)=1$ in this particular case. A continuous random variable random variable has a continuous cumulative distribution function, and in particular $F(x)$ needs to be be continuous where the piecewise densities meet, in this particular case at $x=0$. These together will give you the relevant constants.

For $-1 \le x \le 0$, you want $F(x)=\displaystyle \int_{y=-1}^x (y+1)\,dy$.

For $0 \lt x \le 1$, you want $F(x)=\displaystyle \int_{y=-1}^0 (y+1)\,dy +\int_{y=0}^x (-y+1)\,dy$.

Best Answer

Related Solutions

[Math] Plot the cdf and simulate a random variable (rv) with this cdf using the inversion method.

[Math] Finding the CDF of a piecewise PDF

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