The <conditional density> gives
$$
\mathbb E[Y\mid X=x]=\int_0^\infty yxe^{-xy}\,dy=\frac1x,
$$
using either <integration by parts> or the <expected value> of an <exponential distribution>. Hence
$$
\boxed{\mathbb E[Y\mid X]=\frac1X\quad\text{almost surely}.}
$$
This is a finite value for almost every realization of $X$, but not an integrable <random variable>: $\mathbb E[1/X]=\int_0^\infty e^{-x}/x\,dx=\infty$. Since $Y\geq0$, its <conditional expectation> remains well-defined in the nonnegative, extended-expectation sense; the <tower property of conditional expectation> then gives $\mathbb EY=\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 4F / ii / Solution

Integrate the <joint probability density> over $y$ to obtain the <marginal density> of $X$:
$$
f_X(x)=\int_0^\infty xe^{-x(y+1)}\,dy=e^{-x},\qquad x>0.
$$
Thus $X$ has an <exponential distribution> of rate one. Dividing the <joint probability density> by this <marginal density> gives the <conditional density>
$$
\boxed{f_{Y\mid X=x}(y)=xe^{-xy}\mathbf1_{\{y\geq0\}},\qquad x>0.}
$$
Equivalently, $Y\mid X=x$ has an <exponential distribution> of rate $x$. The value at $x=0$ can be assigned arbitrarily: it is a probability-zero conditioning value and the density ratio there is not meaningful.

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 4F / i / Solution

{title2=$f_{Y\mid X=x}(y)$}

= Conditional probability density
{synonym}

Where the <marginal density> $f_X(x)$ is positive, a <joint probability density> determines the conditional density
$$
f_{Y\mid X=x}(y)=\frac{f_{X,Y}(x,y)}{f_X(x)}.
$$
It integrates to one in $y$, and supplies <conditional expectations> by integrating against it. Its values on a probability-zero set of conditioning points are not fixed by the <joint probability distribution>.

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