0
$\begingroup$

Random variable X follows geometrical distribution with p=1/4. Random variable Y follows uniform distribution in [-X,X]. I'm looking for P(Y>3/2) and also P(X=2|Y>3/2).I know for a fact that Σ(from k=1 to infinity)zk/k=-log(1-z) for |z|<1.

$\endgroup$
0
$\begingroup$

The probability that $Y \gt 3/2$ is actually the probability that $ 3/2 \lt Y \lt X$.

If $X \lt 1$ then it is impossible for $Y$ to be greater than $3/2$.

Therefore $X$ must be at least $2$ for the probability to even make sense. So we have $$\mathbb{P} \left(Y \gt \frac32 \right) = \sum_{x=2}^\infty \left( \frac14 \right)\left( \frac34 \right) ^{x-1} \left( \frac{x - 3/2}{2x}\right) $$

See if you can take it from here

$\endgroup$
  • 1
    $\begingroup$ No, it is possible for $Y>1.5$ when $X=2$ $\endgroup$ – Graham Kemp May 20 at 23:19
  • 1
    $\begingroup$ @Graham Kemp True, I simply misread it as $Y > 3$ when I solved the problem. Will fix $\endgroup$ – WaveX May 20 at 23:20
  • $\begingroup$ I can't make something out of this. $\endgroup$ – Panagiotis Panagiotou May 22 at 17:23
  • $\begingroup$ This sum converges to about $.2159$ $\endgroup$ – WaveX May 22 at 17:39

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.