I'm trying to solve this problem set from my lecture notes in introductory probability theory:
Could you please verify my attempt? Thank you so much!
My attempt:
Let $f$ be the p.m.f of $Y$.
1.
$f(k) = \mathbb{P} (Y=k) = \begin{cases} \mathbb{P} (X= k/2) = {n \choose k/2} p^k (1-p)^k &\text{if} \quad 2|k \\ 0 &\text{otherwise} \end{cases}$
2.
$f(x) = \mathbb{P} (Y=x) = \mathbb{P} (X = \ln x) =\frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{\ln^{2} (x)}{2}\right) = \frac{1}{\sqrt{2 \pi}} x^{\frac{-\ln x}{2}}$.
3.
$f(x) = \mathbb{P} (Y=x) = \mathbb{P} (X = \ln x) = \lambda \exp (-\lambda \ln x) {1}_{\mathbb R_+} (\ln x) = \lambda x^{-\lambda} {1}_{\mathbb R_+} (\ln x) =$ $\lambda x^{-\lambda} {1}_{(1,\infty)} (x)$.
4.
$f(x) = \mathbb{P} (Y=x) = \mathbb{P} (X =\arccos (x)) = \frac{1}{b-a} {1}_{[0,\pi/2]} (\arccos (x))$.
1) is fine. But 2), 3) and 4) are wrong since for continuous random variable, $\mathbb P\{X=x\}=0$ for all $x$. The idea is the following one :
For 2) : if $y\leq 0$, then $\mathbb P\{Y\leq y\}=0$. Suppose $y>0$. Then,
$$\mathbb P\{Y\leq y\}=\mathbb P\{e^X\leq y\}=\mathbb P\{X\leq \ln(y)\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty }^{\ln(y)}e^{-\frac{x^2}{2}}\,\mathrm d x.$$ Therefore, the density function is given by $$f_Y(y)=\begin{cases}0&y\leq 0\\ \frac{1}{y\sqrt{2\pi}}\cdot e^{-\frac{\ln^2(y)}{2}}&y>0\end{cases}.$$ The rest can be made in the same way.