Let X be a random variable that assumes all the non-negative integral values. We are given that for any positive integer:
$k · \mathbb{P}(X = k) = 10 · \mathbb{P}(X = k − 1)$
Find the distribution of X.
Can anyone help me with that?
I started it like that:
$\mathbb{P}(X = k) = 10/k \cdot \mathbb{P}(X = k-1) = 10^2/{k\cdot(k-1)}\cdot \mathbb{P}(X=k-2)=...=\frac{10^{(k-1)}}{{k\cdot(k-1)\cdot ... \cdot(k-(k-1))}}\cdot\mathbb{P}(X=0)$
I know the answer should be that $X$~$Poiss(\lambda = 10)$
but I dont know how to turn it into $P(X=k)=\frac{10^k}{k!} \cdot e^{-10}$
thanks
You are already almost there: observe that $$k(k-1)\cdots(k-(k-1)) = k(k-1)(k-2) \cdots (2)(1) = k!.$$ But you also have a small error, since $$\Pr[X = 1] = \frac{10}{1} \Pr[X = 0], \\ \Pr[X = 2] = \frac{10^2}{2(1)} \Pr[X = 0], \\ \Pr[X = 3] = \frac{10^3}{3(2)(1)} \Pr[X = 0],$$ and so forth. So your formula should be $$\Pr[X = k] = \frac{10^k}{k!} \Pr[X = 0],$$ and not $10^{k-1}$ in the numerator. Now to find $\Pr[X = 0]$, we must have the sum of all probabilities must equal to $1$: $$1 = \sum_{k=0}^\infty \Pr[X = k] = \sum_{k=0}^\infty \frac{10^k}{k!} \Pr[X = 0] = \Pr[X = 0] \sum_{k=0}^\infty \frac{10^k}{k!} = \Pr[X = 0] e^{10},$$ where in the last step we have used the series expansion for the function $e^z$: $$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}.$$ Therefore, $\Pr[X = 0] = e^{-10}$ and the result follows.