My question is based on this post. In summary, $X \sim \text{Unif}(a,b)$ and $Y|X \sim \text{Unif}(a,X)$. Then the author does the following calculations:

\begin{align} f(y) = \int_{-\infty}^{\infty} f(y|x) f(x) dx & = \int_{y}^{b} \frac{1}{x-a} \frac{1}{b-a} dx \\ & = \frac{1}{b-a} \int_{y}^{b} \frac{1}{x-a} dx \\ &= \frac{1}{b-a} \left[ \log(b-a)-\log(y-a) \right] ,\quad a<y<b \end{align}

I'm not sure how the author changes the support from $[a,b]$ to $(a,b)$. The support of $X$ is $[a,b]$ so $Y$ could have support $[a,a]$ and the limits of integration should go from $a$ to $b$. However, this would lead to a divergent integral. In a nutshell, what justifies $a < y < b$ in the last line?

  • 1
    $\begingroup$ A support of $(a,b)$ is equivalent to a support of $[a,b]$ since the endpoints have measure zero against the Lebesgue measure $dx$. The event $X=a$ has probability zero to occur. $\endgroup$ – Xi'an Nov 2 '18 at 4:52
  • $\begingroup$ I see. What's the justification for, in the first line, changing the limits of integration of $(-\infty, \infty)$ to $(y,b)$? $\endgroup$ – user1691278 Nov 2 '18 at 5:19

Both densities involve indicators$$f_X(x)=\mathbb{I}_{(a,b)}(x)\big/(b-a)\quad f_{Y|X}(y|x)=\mathbb{I}_{(a,x)}(y)\big/(x-a)$$and$$f_{X,Y}(x,y)=\mathbb{I}_{(a,b)}(x)\mathbb{I}_{(a,x)}(y)\big/b-y)(x-a)$$This implies $$\mathbb{I}_{(a,b)}(x)\mathbb{I}_{(a,x)}(y)=\mathbb{I}_{(a,b)}(y)\mathbb{I}_{(y,b)}(x)$$hence \begin{align}f_{X|Y}(x|y)&=\mathbb{I}_{(y,b)}\big/(x-a)\,\left\{\int_y^b (x-a)^{-1}\,\text{d}x\right\}^{-1}\\&=\mathbb{I}_{(y,b)}\big/(x-a)\,\{\log(b-a)-\log(y-a)\}^{-1}\end{align}and $$f_Y(y)=\mathbb{I}_{(a,b)}(y)\,\{\log(b-a)-\log(y-a)\}\big/(b-a)$$


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