# find the mean and variance of this poisson random variable

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.

• 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 $X$.) ${}\qquad{}$ – 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, $$\operatorname{E}[X] = \operatorname{Var}[X] = \lambda.$$
• There's also a common factor of $\lambda^2/2$ that cancels out. – Michael Hardy Oct 30 '14 at 1:33
• @user125535 The MathJax code for $\lambda$ is \lambda. – Graham Kemp Oct 30 '14 at 2:45