3

Let $X$ be a random variable with pdf $p(x)$, and let $Y$ be a constant whose pdf can be regarded as $p(y) = \delta(y-Y)$, where $\delta(\cdot)$ is the Dirac-delta function.

Then, $p(x, y) = p(x|y)p(y) = p(x|y)\delta(y-Y) = p(x|Y)\delta(y-Y)$. Since $\int p(x, y) dy = p(x)$, we have $p(x) = \int p(x, y) dy = \int p(x|y)\delta(y-Y) dy = p(x|Y)$, which means $p(x|Y) = p(x)$.

Therefore, we have $p(x, y) = p(x)\delta(y-Y) = p(x)p(y)$, and $X$ and $Y$ are independent.

I am not sure if I am right. If you could give me some references (maybe some books on degenerate distributions), that will be better, and thanks very much in advance!

Graham Kemp
  • 129,094
Ryan
  • 655
  • 1
    What you have is correct. What part are you unsure about? – Paul Sep 08 '16 at 02:10
  • @Paul Thanks very much, and I just want to double check it. – Ryan Sep 08 '16 at 02:11
  • @LeeDavidChungLin Y is a constant, so I think it cannot affect the integral. – Ryan Sep 08 '16 at 02:12
  • @LeeDavidChungLin I am not quite sure what you mean; cf. https://math.stackexchange.com/questions/1585379/independence-between-a-constant-random-variable-and-another-random-variable – kisten Mar 04 '21 at 13:47

1 Answers1

1

Well, $p(x\mid Y)=p(x)$ does not mean $p(x\mid y)=p(x)$.   In fact, $p(x\mid y)$ is undefined when $y\neq Y$ (though it is often convenient to say it is $0$).

Still, that does indicate that: $$p(x,y)=\begin{cases}p(x\mid Y)p(y) &:& y=Y\\ 0 &:& y\neq Y\end{cases} = p(x)p(y)$$

Which is all you need.

Graham Kemp
  • 129,094
  • I agree with you that $p(x|y)$ is undefined for other $y \neq Y$. Thanks very much, Graham! – Ryan Sep 08 '16 at 02:46