Suppose $H$ is a random variable with pdf $f_H(h)$.

Let $X$ and $Y$ be random variables with joint pdf

$$f_{X,Y} = f_H(x) f_H(y)$$

Prove $$P(X \ge Y) = 1/2$$

Is it possible to answer the above basic probability question with a symmetry argument?

In the following advanced probability questions a symmetry argument is not only possible but also necessary because we don't know if the random variables have pdfs:

Prove symmetry of probabilities given random variables are iid and have continuous cdf

Prove independence of events given random variables are iid and have continuous cdf

Confusion with real numbers and random variables; Integration and Independence in Williams

Without symmetry argument:

$$P(X \ge Y) = \int_{\mathbb R} \int_{-\infty}^{x} f_H(x)f_H(y) dx dy$$

Let $$u = \int_{-\infty}^{x} f_H(y) dx$$

to get

$$= \int_0^1 u du = 1/2$$

(Side questions: Do we even need $H$? The fact that $X$ and $Y$ have a joint pdf that splits up into a function of $x$ and a function of $Y$ means that the joint pdf splits up into the pdfs of $X$ and $Y$ assuming they exist...do they?)

That's allowed I'm guessing because even if somehow the random variables don't have pdfs (which they always do anyway in basic probability), $X$ and $Y$ together have a joint pdf, which is all we need.

But is there a way to do this in a symmetric way?

It seems that $X$ and $Y$ are independent because the joint pdf splits up into independent (not in the probability sense) functions of $x$ and $y$:

$$f_X(x) = kf_H(x), f_Y(y) = \frac{1}{k}f_H(y)$$

Am I allowed to say the following?

$$f_X(x) = kf_H(x), f_Y(y) = \frac{1}{k}f_H(y)$$

$$\color{red}{\stackrel{?}{\to}} kf_H(x) = \frac{1}{k}f_H(x)$$

$$\to k = 1$$

$$\to f_X(x) = f_Y(x)$$

I guess that would make sense if $X$ and $Y$ had the same range or something?

Assuming that is allowed, it appears $X$ and $Y$ have the same distribution.

Is there a way to argue that

$$P(X = \max\{X,Y\}) = P(Y = \max\{X,Y\})$$

using basic probability and without integration?

  • 1
    $\begingroup$ I would have thought you could say (a) they are independent since you can separate the densities and (b) they are identically distributed; so by symmetry $P(X \gt Y)= P(Y \gt X)$. Combine this with (c) $P(X=Y)=0$ since they have a density, and you get your result $\endgroup$ – Henry Jul 11 '16 at 23:54
  • $\begingroup$ @Henry How do you know they are identically distributed? How do know that $$P(X > Y) = P(Y > X)$$? What do you mean by 'symmetry' (in a basic probability sense) ? $\endgroup$ – BCLC Jul 11 '16 at 23:55
  • 2
    $\begingroup$ If we do not have prior machinery set up, the symmetry argument takes longer than the integration argument. $\endgroup$ – André Nicolas Jul 12 '16 at 0:02
  • $\begingroup$ @AndréNicolas Ah, thanks. Answers to other questions then please? $\endgroup$ – BCLC Jul 12 '16 at 0:06
  • 1
    $\begingroup$ @BCLC: You proof that under the given conditions $X$ and $Y$ have the same distribution is correct. Maybe one could assert this is the case by symmetry. But there seems no way of avoiding (unless one knows it already) proving that $\Pr(X=Y)=0$. $\endgroup$ – André Nicolas Jul 12 '16 at 0:10

We have the joint probability density function: $f_{X,Y}(x,y) := f_H(x)f_H(y)$ where $f_H$ is itself a probability density function of a continuous random variable.

Then we can find the marginals are both $f_H$. $$\begin{align}f_X(x) =&~ \int_\Bbb R f_H(x)f_H(t)\operatorname d t \\[1ex]=&~ f_H(x) \int_\Bbb R f_H(t)\operatorname d t \\[1ex] =&~ f_H(x)\\[2ex]f_Y(y) = &~ f_H(y) & \textsf{similarly} \\[2ex]\therefore~ f_{X,Y}(x,y) =& ~ f_X(x)\cdot f_Y(y) \end{align}$$

Hence these two random variables have identical, independent, and continuous distributions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.