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.
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, $$\operatorname{E}[X] = \operatorname{Var}[X] = \lambda.$$