Let X be a Poisson random variables with $\lambda = 3.$ Find $E(X\mid X>2).$

1.5k Views Asked by At

So I write this out as,

$$E(X\mid X>2) = \frac 1 {P(X>2)}\sum_{k=2}^\infty k\frac{3^k e^{-3}}{k!}$$

For $P(X>2)$ I get .80085 and when I multiply by the sum I get 3.55947. My book, however, says that the answer is 4.16525.

PS, I computed $P(X>2)$ by subtracting $P(X=0) + P(X=1)$ from $1.$ I hope I didn't mess this up somehow so I assume I made a mistake in computing the summation.

Thank you in advance. Pardon any mistakes in my math. Hopefully it's not something too silly :P

1

There are 1 best solutions below

1
On BEST ANSWER

You are asked to require $X>2$, so $2$ itself is not included. Your computation correctly gives $E[X \mid X \geq 2]$.