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Where the marginal density $f_X(x)$ is positive, a joint probability density determines the conditional densityIt 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.

For a fixed $x\in [0,1]$, the allowed $y$ interval is $[0,\sqrt{x}]$. Integrating the joint probability density over it gives the marginal densityFor a fixed $y\in [0,1]$, the inequality $y\le \sqrt{x}$ is equivalent to $x\ge y^2$. The other marginal density is thereforeAs a normalization check, ${\int }_0^{1}3x^{2}dx=1$ and ${\int }_0^{1}3y(1-y^4)dy=1$.

Take $\lambda >0$ as the rate of the exponential distribution. By independence, the joint probability density of the original pair is ${\lambda }^2{e}^{-\lambda (x_1+x_2)}$ on $x_1,x_2>0$. Put $s=Y_1$, $r=Y_2$. The inverse change of variables isIts Jacobian determinant isUsing the absolute value in the change of variables formula, the transformed joint probability density isand zero otherwise. The support is the entire positive quadrant, and the factors are separately normalized:Integrating out either variable therefore yields these factors as the marginal densities. Factorization of the joint probability density proves independence, withThe sum has a shape-two gamma distribution, and the ratio has a beta-prime distribution with parameters one and one, independent of $\lambda$. Its cumulative distribution function is $r/(1+r)$ for $r>0$.

Integrate the joint probability density over $y$ to obtain the marginal density of $X$:Thus $X$ has an exponential distribution of rate one. Dividing the joint probability density by this marginal density gives the conditional densityEquivalently, $Y\mid X=x$ 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.

Obtain each marginal density by integrating the joint probability density function over the other variable:Each marginal density is zero off its displayed interval and has total integral one.

The support condition $-y<x<y$ forces $y>0$. As a check, the joint probability density integrates toFor any real $x$, the marginal density of $X$ isDividing the joint probability density by this marginal density gives the conditional density:Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two exponential distribution.

For $U=X^2$, use the two inverse branches $x=\sqrt{u}$ and $x=-\sqrt{u}$. Each branch has absolute Jacobian determinant $1/(2\sqrt{u})$. Both lie in the original support exactly when $u>0$ and $y>\sqrt{u}$. The change of variables for probability densities therefore yieldsNo atom of a measure is created at $u=0$, since $\mathbb{P}(X=0)=0$. The value of the probability density function at this boundary may be assigned arbitrarily. Checking the integral over $0<u<y^2$ gives $4y{e}^{-2y}$, which integrates to one.