# Exercise on finding the joint probability distribution

This Exercise is particularly important to me because so far I believe to have a rather poor understanding on how to compute the joint probability distribution.

Problem: Let $X$ be a RV with density $f(x)= \frac{1}{\pi\sqrt{x(1-x)}}$ for $x \in (0,1)$ and $Y$ be a RV with Exponential Distribution (standard, parameter 1). Assuming that $X,Y$ are independent I am supposed to find the joint distribution of $U=XY, \ V=(1-X)Y$

My approach: Although I couldn't rigorously proof it I think I can state that $(X,Y)$ has joint probability density given by $m(x,y)= \frac{1}{\pi \sqrt{x(1-x)}}e^{-y}$ for $(x,y) \in (0,1) \times (0, \infty)$

My idea was now to compute for an arbitrary bounded continuous $g:\mathbb{R^2} \to \mathbb{R}$ $$E(g(U,V)) = \int_{\mathbb{R^2}}g(u,v) h(u,v)dudv$$ and hope that I can find a density function $h$.

$$E(g(U,V))=E(g(XY,(1-X)Y)) \\ \overset{3)}= \int_{\mathbb{R}^2}g(xy,(1-x)y) \frac{1}{\pi \sqrt{x(1-x)}}e^{-y}1_{(x,y) \in (0,1) \times (0 , \infty)} dx dy \\ = \int_{(0,1) \times (0, \infty)} g(xy,(1-x)y) \frac{1}{\pi \sqrt{x(1-x)}}e^{-y} dx dy =:I$$

Choosing the obvious transformation/substitution $(u,v)=(xy,(1-x)y)$ I get $(u,v) \in (0, \infty) \times (0, \infty)$ and $x= \frac{u}{u+v},y=u+v$ for the Jacobi Matrix I obtain $$J= \begin{pmatrix} \frac{v}{(u+v)^2} & \frac{-u}{(u+v)^2} \\ 1 & 1 \end{pmatrix} \implies |\det J| = \frac{1}{u+v}>0$$ So finally I would obtain for the above integral denoted as $I$ that $$I= \int_{(0, \infty)^2} g(u,v) \underbrace{\frac{1}{\pi(\sqrt{\frac{u}{u+v}(1-\frac{u}{u+v})}}e^{-(u+v)} \frac{1}{u+v}}dudv \\ = \int_{(0, \infty)^2} g(u.v) \frac{1}{\pi \sqrt{uv}} e^{-(u+v)}dudv$$

Questions: 1) The obvious question of course if the above is correct or not

2) Do I need to do anything more? Or just state that the distribution of $(U,V)$ is given by the strange underbraced term on the last integral?

3) Given the density function of $(X,Y)$ (assuming my formula in the first paragraph is correct) why is this equation true? Intuitively I don't see why this should hold given the more standard formula $$E(f(X))= \int_{\mathbb{R}^d} f(x) P_X(dx)$$

Additional Question (optional): I am very new to this topic and have little to no to experience, please if you know of a more elegant way to approach the solution I would gladly know about it.

• Update: I think my integration limits are wrong because I said $v=(1-x)y$ where $x \in (0,1)$ and $y \in (0, \infty)$ it follows that $v \in (0, \infty)$ which is good, because that makes the Determinant of the Integral much better defined (no division by zero possible). I will update that. – Spaced Apr 23 '16 at 15:16
• (1) Because both $u$ and $v$ are positive, you can substantially simplify the "strange underbraced term". (2) In question 3, what do you mean by "this equation"? (3) The calculus of differential forms is arguably "more elegant." It's certainly simpler! – whuber Apr 23 '16 at 16:49
• Thanks a lot @whuber, I have done further simplifications. For my question 3 I highlighted it in the above question, meaning the 3) above the = sign. I wondered because I do understand that $$E(f((X,Y)))= \int f(x,y)h(x,y)dxdy$$ where $h$ is a density function of $(X,Y)$. However I do not understand why I can still use the same density function if I "change" $X,Y$ slightly for example I used something in the nature $$E(f(X+Y,XY))=\int f(x+y,xy)h(x,y)dxdy$$ i.e. I did not change the density function at all. Since you didn't comment on it, I suppose my calculations are indeed correct? – Spaced Apr 23 '16 at 17:13
• Conclusion: $(U,V)$ has density given by $1_{(u,v) \in (0, \infty )^2} \frac{1}{\pi \sqrt{ uv}}e^{-(u+v)}$, additionally the random variables $U,V$ are independent because if I denote $$h(u,v)=1_{(u,v) \in (0, \infty )^2} \frac{1}{\pi \sqrt{ uv}}e^{-(u+v)}$$ then $h(u,v)=h_1(u)h_2(v)$ where $$h_1(u)=1_{u \in (0, \infty)} \frac{1}{\sqrt{\pi u}} e^{-u}, \ h_2(v)= 1_{v \in (0, \infty)} \frac{1}{\sqrt{ \pi v }}e^{-v}$$ and $h_1, h_2$ are densities on $U$ respectively $V$. – Spaced Apr 23 '16 at 17:30

Statistical reasoning provides an elegant solution.

Because the integral of $f$ is used to define inverse trig functions, one is immediately tempted to interpret $X=\sin^2(A)$ for a random variable $A$ ranging from (say) $0$ to $\pi/2$. Substituting $\sin(a)$ for $x$ in $f$ gives

$$f(x)\,\mathrm{d}x = f(\sin^2(a))\mathrm{d}\left(\sin^2(a)\right) = \frac{2\sin(a)\cos(a)\,\mathrm{d}a}{\pi\sqrt{\sin^2(a)(1-\sin^2(a))}}=\frac{2}{\pi}\mathrm{d}a.$$

This reveals $X$ as the squared sine of a uniformly distributed angle on $[0,\pi/2)$. Consequently $1-X=\cos^2(A)$ is its squared cosine.

Recall (this is familiar from the study of the Normal distribution and related distributions of statistical importance) that an Exponential variable $Y$ has the same distribution as half the sum of squares of two independent standard Normal variables $Z_1$ and $Z_2$. In the plane, the ordered pair $\mathbf{Z}=(Z_1,Z_2)$ has a standard bivariate Normal distribution, showing that $Y$ is half the squared length of $\mathbf{Z}$, $$Y=1/2\,|\mathbf{Z}|^2.$$

Consequently

$$U=XY = 1/2\,\sin^2(A)|\mathbf{Z}|^2 = 1/2\,\left(\sin(a)|\mathbf{Z}|\right)^2$$

and

$$V=(1-X)Y = 1/2\,\cos^2(A)|\mathbf{Z}|^2 = 1/2\,\left(\cos(a)|\mathbf{Z}|\right)^2.$$

Those expressions that have been squared are the very components of $\mathbf{Z}$ itself:

$$U = 1/2\,Z_1^2,\ V=1/2\,Z_2^2.$$

Apparently $U$ and $V$ are independent and their distributions are both--by definition--half a $\chi^2(1)$ distribution. It is now easy to write down their joint distribution any way you wish: as a PDF, CDF, characteristic function, moment-generating function, cumulant-generating function, etc. But it's probably most revealing to have expressed them in this familiar statistical form.

A quick simulation supports these conclusions: by simulating $U$ and $V$ independently as proportional to $\chi^2(1)$ variates and solving

$$Y=U+V;\ X=U/Y$$

we can see whether $X$ and $Y$ have the distributions originally assumed of them. A quick check--which could be formally verified with a goodness of fit test (like a chi-squared test)--is to overplot the histograms of the simulated $X$ and $Y$ with the density functions. They should match, up to a small amount of random variation in the areas of the histogram bars. They do.

Here is the R code that made this figure.

n <- 1e5
set.seed(17)

u <- 1/2 * rchisq(n, df=1)
v <- 1/2 * rchisq(n, df=1)
y <- u + v
x <- u / y

par(mfrow=c(1,2))
hist(x, freq=FALSE)
curve(1 / (pi * sqrt(x*(1-x))), col="Red", lwd=2, add=TRUE)
hist(y, freq=FALSE)