Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

x, y are independent possion variates. variance of X+Y = 9 P(X = 3/X+Y=6) = 5/54

Can anyone help me find the mean of X?

Does Chebyshev's inequality come into picture?

share|cite|improve this question

I don't see what you want to do with Chebyshev's inequality.

Let $\lambda_X$ be the parameter for $X$ and $\lambda_Y$ be the parameter for Y.

Then $$9 = Var[X+Y] = Var[X] + Var[Y] = \lambda_X + \lambda_Y$$

and $$P(X = 3 | X+Y = 6) = P(Y = 3) = \frac{\lambda_Y^3 e^{-\lambda_Y}}{6} = \frac{5}{54}$$

This isn't a nice-looking system but you should be able to solve for the mean $\lambda_X$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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