probability distributions
Let $X$ be a standard exponential random variable, and $Y = \log X$.
(a) Find DIRECTLY the c.d.f of $Y$ and use it to calculate the density of $Y$.
(b) Find DIRECTLY the p.d.f of $Y$.
So far, I did:
(a) $F_Y(y) = P(\log X \leq y) = P(X \leq e^y) = F_X(e^y)
f_Y(y) = F'_Y(y) = F'_X(e^y) = [f_X(e^y)][e^y] = (e^{-e^y})(e^y) = e^{y-e^y}$
(b) $f_Y(y) = P(\log X = y) = P(X = e^y) = f_X(e^y) = e^{-e^y}$
How come the answers I got for part a and b are not the same. What did I do wrong?
The first thing to do in such a problem is to figure out the support of the random variable $Y$, which is $(1, \infty).$ To find it, try several values of $x$ in $(0, 1)$ and see what happens. You were right to ask about this crucial issue.
Your text probably shows transformation methods based on the CDF and on the PDF. An abbreviated version of the former (with gaps for you to complete) is as follows:
$$ F_Y(t) = P(Y \le t) = P(1/X \le t) = P(X \ge 1/t)\\ = 1 - P(X \le 1/t) = 1 - 1/t^2,$$
for $t$ in the support of $Y$.
Then, by differentiation, the PDF of $Y$ is $f_Y(t) = 2t^{-3},$ as you say. Finally, $E(Y) = \int_1^\infty 2t^{-3} \, dt = 2.$ [Some of this is in your Question, and some is in @drhab's succinct Answer (+1)]
If you recognize that $X \sim Beta(2, 1),$ then it is easy to do a quick simulation in R to check this answer (correct to two or three decimal places based on a million simulated values of $Y$):
x = rbeta(10^6, 2, 1); y = 1/x; mean(y)
## 2.001900 # aprx of E(Y)
Below is a histogram of 10,000 simulated values of $Y$ along with a plot of the PDF of $Y$. This distribution is extremely right-skewed, with a long tail to the right. The scale of the histogram extends out to about 60 in recognition of occasional scattered points too sparsely spread to make histogram bars of noticeable height. (The biggest of the 10,000 values plotted in this histogram was at 57.44494.)
Yes, and you can simply use the transformation formula to find directly the density of $Y$. $g(X) = \sqrt{X}$ is monotone (increasing) function on $[0,3]$, and $g^{-1}(Y) = Y^2$, thus $$ f_Y(y) = f_Y(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right| = \frac{2}{3}y, \quad y\in (0,\sqrt{3}). $$
Recall that the density function is the derivative of the cumulative distribution function, i.e., $$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}P(Y\le y), $$ in your case you can verify that $$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}\frac{y^2}{3} =\frac{2}{3}y. $$
Best Answer
Because PDFs don't work like this. The issue is you need to change of variables on the probability $P(Y=y)=f_Y(y)dy$, and not just the density. More precisely, if $y=\log(x)$ then $dy=e^{-y}dx$, so $P(Y=y)=f_Y(y)dy=P(\log X=y)=f_X(e^y)dx=f_X(e^y)e^ydy$ so that you recover your answer from the first part.