0
$\begingroup$

For the following experiment:

A random number $X$ is chosen uniformly from $[0, 1]$. Then a sequence $Y_1,Y_2\dots Y_i$ of random numbers is chosen independently and uniformly from $[0, 1]$. The game ends the first time that $Y_i > X$.

If I define $Z$ as a discrete (countably infinite) random variable $\{1,2,3...\}$ that represents the number of turns before the game ends (inclusive of the last turn), $P(Z=z|X=x)=(1-x)x^{z-1}$.

  1. The marginal probability mass function for $Z$ is: $p_Z(z)=\int_{0}^{1}P(Z=z|X=x)f_X(x)dx=\int_{0}^{1}(1-x)x^{z-1}dx$?

  2. Is $E(Z|X=x) = \frac{1}{1-x}$, and hence $E(Z) = \int_{0}^{1}E(Z|X=x)f_X(x)dx=\int_{0}^{1}\frac{1}{1-x}dx$ which diverges?

  3. How is $f_{X,Z}(x,z)$ defined, given that it is neither continuous nor discrete?

$\endgroup$
  • $\begingroup$ 1. and 2. look correct. For 3. to get the z component of the density function you need to use delta functions at each value of z. $\endgroup$ – herb steinberg May 27 at 3:00
0
$\begingroup$

The joint distribution-mass distribution function of $\ X\ $ and $\ Z\ $ is given by $$ P\left(X\le x, Z=z\right) =\cases{ 0 & for $x<0 $ \\ \int_\limits{0}^x \left(1-y\right)y^{z-1}dy=\frac{x^z}{z}-\frac{x^{z+1}}{z+1} & for $0\le x<1 $\\ \frac{1}{z(z+1)} &for $1\le x $}\ ,$$ and their joint density-mass distribution function $\ f_{X,Z}\ $ (assuming that's what you meant by this expression) is obtained by differentiating this with respect to $\ x\ $: $$ f_{X,Z}\left(x,z\right)=\cases{ 0& for $x<0$\\ x^z-x^{z+1} & for $0\le x<1 $\\ 0 & for $1\le x $} $$

Calculating $\ E\left(Z\right)\ $ directly: $$\sum_\limits{z=1}^\infty zp_Z\left(z\right)=\sum_\limits{z=1}^\infty z\left( \frac{1}{z(z+1)}\right)=\sum_\limits{z=1}^\infty\frac{1}{z+1}$$ confirms your conclusion that it diverges (to $+\infty$).

$\endgroup$

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.