# Find the probability density functions of $X$ and $Y$

If $X$ and $Y$ have joint density function

$f_{X, Y}(x, y)= \begin{cases} 1/y, &\text{if } 0<y<1, &0<x<y \\ 0, &\text{otherwise} \end{cases}$

find the probability density function of $X$, i.e., $f_X(x)$ and the probability density function of $Y$, i.e., $f_Y(y)$.

My attempt:

The formulas for $f_X(x)$ and $f_Y(y)$ are:

$$f_X(x)=\int_{-\infty}^\infty f_{X, Y}(x, y)\ dy$$

$$f_Y(y)=\int_{-\infty}^\infty f_{X, Y}(x, y)\ dx$$

I can substitute $1/y$ for $f_{X, Y}(x, y)$ but I'm unsure what the new limits of integration would be. Any suggestions?

## 1 Answer

Here's an attempt.

We have the upper triangular region of the $[0,1] \times [0,1]$ as our domain of integration, so we just need to figure out how to write these bounds.

We have $\int_0^1 \int_0^y 1/y \text{ } dx \text{ } dy = \int_0^1 \int_x^1 1/y \text{ } dy \text{ } dx = 1$, so the marginal $f_Y(y)$ can be found as $$\int_0^y f_{X,Y}(x,y) dx = 1$$ and similarly the marginal $f_X(x)$ can be found as $$\int_x^1 f_{X,Y}(x,y) dy = -\log(x)$$

I also wrote a small Metropolis-Hastings MCMC program to simulate from this distribution as a sanity check and have included my R code below.

fxy <- function(x,y){
1/y * (x<y) * (y<1) * (x>0) * (y>0)
}

mh <- function(niter,xstart,ystart,propsd){
x=rep(NA,niter)
y=rep(NA,niter)
res = list(x=x,y=y)
res$x = xstart res$y = ystart
for(i in 2:niter){
xold <- res$x[i-1] yold <- res$y[i-1]
xnew <- xold + rnorm(1,0,propsd)
ynew <- yold + rnorm(1,0,propsd)
A <- min(1,fxy(xnew,ynew)/fxy(xold,yold))
if(runif(1) < A){
res$x[i] <- xnew res$y[i] <- ynew
} else {
res$x[i] <- xold res$y[i] <- yold
}
}
return(res)
}

niter <- 1e4
res <- mh(niter=niter,xstart=0.25,ystart=0.5,propsd=0.5)
hist(res$x,freq=F) xx <- seq(0.01,1,by=0.01) lines(-log(xx)~xx,col='red') hist(res$y)


Plotting the histograms we see $f_Y(y) \propto 1$ and $f_X(x) \propto -\log(x)$  