Q. Let $X$ be a random variable with the probability density function
$
f_{X}(x) = \begin{cases} 1 &\text{ if} \quad 0 < x < 1 \\
0 &\text{ otherwise}
\end{cases}
$
Let $Y$ be a random variable with the conditional probability density function
$
f_{Y|X}(y|x) = \begin{cases}
1/x &\text{ if} \quad 0 < y < x \\
0 &\text{ otherwise}
\end{cases}
$
What is the marginal probability density function for $Y$?
I see the answer must be $f_{Y}(y) = 2(1 – y)$ for $0 < y < 1$. (Edit: see below.) However, I am having trouble deriving this result. My approach has been to apply the law of total probability in the form
\begin{align*}
f_{Y}(y) &= \int\limits_{-\infty}^{+\infty} f_{Y|X}(y|x) f_{X}(x) \text{ d}x \\
&= \int\limits_{0}^{1} f_{Y|X}(y|x) \text{ d} x \\
&= \int\limits_{0}^{x} \frac{\text{d} x}{x}
\end{align*}
But here I run into the difficulty that $x$ appears in the limit. Is there another way of going about this problem? There is likely something obvious I am missing. Thanks in advance for your help.
UPDATE. Starting with J.G.'s advice on the limits, I found the easier way to approach the problem is first to write the joint density
$
f_{X,Y}(x,y) = \begin{cases}
1/x &\quad \text{if } 0 < y < x < 1 \\
0 &\quad \text{otherwise}
\end{cases}
$
Then the marginal density for $Y$ is
\begin{align*}
f_{Y}(y) &= \int\limits_{-\infty}^{+\infty} f_{X,Y}(x, y) \text{ d} x \\
&= \int\limits_{y}^{1} \frac{\text{ d} x}{x} \\
&= – \ln (y)
\end{align*}
Pretty weird result. Turns out my earlier intuition that $f_{Y}(y) = 2 (1 – y)$ was wrong.
Best Answer
The last integral's lower limit should be $y$, not $0$. Similarly, the upper limit should be $1$.