# joint probability distribution with a constant

If $X \sim N(0,1)$, then what is the joint probability distribution of $(X+1,X)$?

An attempt: $f(x,x+1)=f(x|x+1)f(x+1)=f(x+1)$, so $N((0,0),(0,0;0,1))$. Note sure though...

First, define $Y=X+1$ to make the notation easier. What you are looking for is the pdf of YX. Here it is:

$$f_{YX}(y,x) = f_{Y|X}(y|x)f_X(x).$$

We know $f_X(x)$, so the question is what is $f_{Y|X}(y|x)$ ?

The answer is the following. If we know that $X=x$, then $Y=X+1 = x+1$. So $Y$ is a constant, with pdf given by $\delta(y-(x+1))$. Note that this is Dirac's delta.

The overall pdf:

$$f_{YX}(y,x) = \delta(y-(x+1)) \frac{1}{\sqrt{2\pi}}\exp(-x^2/2).$$

In other words, the pdf is zero whenever $y\neq x+1$. (You could also say that for $y=x+1$ it is equal to "infinity", but this is not very informative).

You can verify this by marginalizing w.r.t. either $X$ or $Y$, to see that you get the correct pdfs.