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.


The last integral's lower limit should be $y$, not $0$. Similarly, the upper limit should be $1$.

  • $\begingroup$ Thanks, that was what I needed to see. $\endgroup$ – A. B. H. Dec 2 '18 at 1:51

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