Let $X$ be the poisson random variable such that $P(X = 2) = 9P(X=4) + 90P(X=6)$

find the mean and variance of $X$.

I'm not sure how to approach this problem..am i supposed to multiply each probability by their respective x value and then add them all together? or am i supposed to somehow find out the values of the probabilities first? Not sure how to find the values with just that equation.

  • $\begingroup$ Don't alternate between capital $X$ and lower-case $X$ like that. Case-sensitivity is standard in mathematical notation, and without it we would not be able to understand the meaning of something like $\Pr(X=x)$ or $\Pr(X\le x)$. (I've changed them all to capital $X$.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 30 '14 at 1:34

Recall that the probability mass function for a Poisson random variable is $$\Pr[X = k] = e^{-\lambda} \frac{\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots.$$ Thus the given condition is equivalent to $$e^{-\lambda} \frac{\lambda^2}{2!} = 9 e^{-\lambda} \frac{\lambda^4}{4!} + 90 e^{-\lambda} \frac{\lambda^6}{6!}.$$ Note that there is a common factor of $e^{-\lambda}$ which cancels out; can you solve the remaining equation for the rate parameter $\lambda$? Then recall that for a Poisson random variable, $$\operatorname{E}[X] = \operatorname{Var}[X] = \lambda.$$

  • $\begingroup$ There's also a common factor of $\lambda^2/2$ that cancels out. $\endgroup$ – Michael Hardy Oct 30 '14 at 1:33
  • $\begingroup$ so..after that i got 960 = 720x^2 + 24x^4.....am i supposed to use quartic formula for this or what?..did i most likely mess up in solving this equation? (x being lambda sorry couldn't remember the lambda mathjex code) $\endgroup$ – user125535 Oct 30 '14 at 1:57
  • $\begingroup$ @user125535 The MathJax code for $\lambda$ is \lambda. $\endgroup$ – Graham Kemp Oct 30 '14 at 2:45

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.