# Joint probability density function using the same random variable

My original problem statement:
Let $x$ be a random variable with pdf $f_x(x)$ and let y be $y=x$.
(a) Find the joint pdf $f_{x,y}(x,y)$.
(b) Find the conditional pdfs $f_{x|y}(x|y), f_{y|x}(y|x)$

I've already given some thoughts how the joint pdf should behave: Picturing the joint pdf as $f_{x,y} = f{y|x}*f(x)$ and assuming a fixed $x$ the result should be 0 for $x \neq y$ and some other value for $x = y$. But I have really a hard time getting the idea which value that might be.

Any hint how this problem can be approached is highly appreciated.

• Distinguish between the statements "$y$ is identically distributed with $x$" and "$y$ is $x$". Yours is the second case. So what could "the joint density of a random variable with itself" mean? – Alecos Papadopoulos Dec 1 '13 at 17:14

The random couple $(X,Y)$ has no PDF since its support is included in the diagonal $\{(x,x)\mid x\in\mathbb R\}$, which has Lebesgue measure zero.
Likewise, the conditional distribution of $X$ conditionally on $Y$ (or vice versa) has no density. For every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is $\delta_y$ the Dirac mass at $y$. This measure has no density with respect to the Lebesgue measure.
Edit: One might add that the only reason why distributions are useful is to compute means. That is, the distribution of a random element $Z$ allows to compute $E[u(Z)]$ for every measurable real-valued function such that the expectation exists. Here, the distribution of $(X,Y)$ is crystal clear since, considering the distribution $P_X$ of $X$, one has $$E[u(X,Y)]=\int u(x,x)\mathrm dP_X(x).$$ For example, if $P_X$ has a density $f_X$, $$E[u(X,Y)]=\int u(x,x)f_X(x)\mathrm dx.$$
• Thanks for the explanation, I understand that my support is a subset of the diagonal $\{(x,x)|x \in \mathbb R \}$. The Lebesgue measure was not introduced to me yet. From wikipedia I have learned that it can be seen as the n-dimensional volume. Can you give me a hint/reason why the volume must be 0 and hence no valid pdf can be found? – daFritz Dec 1 '13 at 17:21
• The volume of any segment in $\mathbb R^2$ is zero and every line is a countable union of such segments. – Did Dec 1 '13 at 20:05