Let $Z\sim N(0,1)$, $X=Z^2$, $Y=e^{-X}$. Compute the expectations to confirm change of variables.

Question

Let $Z\sim N(0,1)$, $X=Z^2$, $Y=e^{-X}$.

There's a couple of parts to this question; the first part I believe is setting up to the part that I posted.

The first part says to find the distributions $P_X$ and $P_Y$ on $(\mathbb{R},\mathcal{B}(\mathbb{R})).$

Letting $\phi(x)$ be the density for the normal distribution, I calculated the following;

$$P_X(X(\omega)\leq x)=P(Z(\omega)^2\leq x) = P(-\sqrt{x}\leq Z(\omega)\leq \sqrt{x}) = \int_{-\sqrt{x}}^\sqrt{x}\phi(t)dt,$$

And I could similarly find $$P_Y(Y(\omega)\leq y)=\int_{-\sqrt{\ln(y)}}^\sqrt{\ln(y)}\phi(t)dt.$$

After this, I'm then asked to calculate $$\int_\mathbb{R}e^{-z^2}\phi(z)dz, \int_\mathbb{R}e^{-x}dP_X, \text{ and } \int_{\mathbb{R}}ydP_Y,$$ and verify that all integrals agree.

Getting the first integral wasn't too hard.

Here's where I'm having trouble: even though I think what I have for the first part is technically correct, it's not very useful in actually calculating the integrals. For example, The only thing I can think to do to calculate $\int_\mathbb{R}e^{-x}dP_X$ would be to find a sequence of simple functions converging to $e^{-x}$ and then taking the limit of the lebesgue integrals w.r.t. $P_X$, which I'm pretty sure is not what is expected in this problem.

More consequentially, this problem is prefaced with the statement that, when using change of variables, one of the three integrals is usually easier to calculate than the others, and tells us ahead of time that the first integral will be easier to get than the others. This leads me to believe that we're expected to calculate the integrals independently, without assuming validity of the CoV formula, in order to demonstrate the usefulness of it.

Is there a non-ridiculous way to get these integrals without assuming change of variables, or do you think I'm misunderstanding something about what this problem is asking of me?

Thanks in advance for any thoughts.

Answer 1

Going on with your calculations (derivating the CDF you found) you will easy realize that

$$X\sim \chi_{(1)}^2$$

$$f_X(x)=\frac{1}{2\sqrt{x}}\phi(\sqrt{x})+\frac{1}{2\sqrt{x}}\phi(-\sqrt{x}) =\dots=\frac{\Big(\frac{1}{2}\Big)^\frac{1}{2}}{\Gamma\Big(\frac{1}{2}\Big)}x^{-\frac{1}{2}}e^{-\frac{x}{2}}$$

For $Y=e^X$,

In this case with monotonic transformation it is faster to get immediately the density with the fundamental transformation Theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

Getting

$$f_Y(y)=\frac{1}{y\sqrt{2\pi logy}}e^{-\frac{logy}{2}}\cdot\mathbb{1}_{(1;\infty)}(y)$$