Prove or disprove the following with reasons:

  1. If $X,Y,Z$ are iid standard exponential random variables then $X + Y$ and $Y + Z$ are both gamma random variables.
  2. If $X,Y$ are continuous random variables with joint density $f(x,y)=2, x, y \ge 0, x+y \le 1$ then X,Y are marignally both $\operatorname{U}[0,1]$
  3. Suppose $X,Y$ have a bivariate uniform distribution in the unit circle. Therefore, $E(XY-XY^2)=0$
  4. If X is a standard exponential random variable then $P(X$ is an even integer$) = 0$.

What I have so far:

  1. True, because all exponential random variables are gamma random variables with $\alpha=1$
  2. False, because $f_X(x)=\int_{0}^{1-x}2dy \ne 1$ (not sure about this one)

Is many answer and reasoning for 1 correct? What about 2 (could someone explain this a bit more since I'm kinda confused)? For 3 and 4 I can only guess an anwer.

  1. What you state is true, but it is not obvious why it has anything to do with the question. If you are citing some result like "the sum of independent gamma random variables with the same $\beta$ parameter is also a gamma random variable" you should make that clear.

  2. Your solution is fine.

  3. Hint: $E[XY-XY^2] = E[(Y-Y^2)E[X \mid Y]]$. Alternatively, you can directly compute the integral $\iint_D (xy-xy^2) \frac{1}{\pi} \mathop{dx} \mathop{dy}$.

  4. Hints: What is $P(X=2)$? What about $P(X=4)$? There is an axiom of probability that allows you to write $P(\text{$X$ is even integer})$ as a sum of many probabilities.

  • $\begingroup$ What does D represent in 3? What are the limits in that integral? Are they $\int_0^1 \int_0^{\sqrt{1-y^2}}$? $\endgroup$ – SomebodyOnEarth Dec 11 '17 at 6:54
  • 1
    $\begingroup$ @SomebodyOnEarth I just meant integral over the unit circle. I think the limits should be $\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}$. $\endgroup$ – angryavian Dec 11 '17 at 7:04
  • $\begingroup$ So 3 is true and 4 is false, correct? $\endgroup$ – SomebodyOnEarth Dec 11 '17 at 7:06
  • 1
    $\begingroup$ (4) is true. The exponential distribution is continuous. Therefore, it puts probability 0 on every individual point, including all the integers. $\endgroup$ – BruceET Dec 11 '17 at 7:34

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