Find the probability that Y<= 1?

Hello,

Quote:

Originally Posted by Yan

Suppose Y is uniform on [0;X + 1], where X is Poisson with mean 3. Find the
probability that Y <= 1.

Can you show all your work? thanks

$\mathbb{P}(Y\leq 1)=\sum_{k=0}^\infty \mathbb{P}(Y\leq 1 \mid X=k)\mathbb{P}(X=k)=\sum_{k=0}^\infty \mathbb{P}(Y\leq 1 \mid X=k) \cdot\frac{3^k e^{-3}}{k!}$

But then, which distribution follows the random variable $Y\mid X=k$ ? It's a uniform distribution over $[0,k+1]$
Where the pdf is $f(y)=\begin{cases}\frac{1}{k+1} & \text{ if } y\in[0,k+1] \\ 0 & \text{ otherwise} \end{cases}=\frac{1}{k+1} \cdot \bold{1}_{[0,k+1]}(y)$

$\mathbb{P}(Y\leq 1\mid X=k)=\int_0^1 \frac{1}{k+1} ~dy$ (because $k+1\geq 1$ almost surely, we get these boundaries)

Which is $\mathbb{P}(Y\leq 1\mid X=k)=\frac{1}{k+1}$

Hence
$\begin{aligned} \mathbb{P}(Y\leq 1) &=\sum_{k=0}^\infty \frac{3^k e^{-3}}{(k+1)!} \\ &=\frac 13 \sum_{k=0}^\infty \frac{3^{k+1}e^{-3}}{(k+1)!} \\ &=\frac 13 \sum_{k=1}^\infty \frac{3^k e^{-3}}{k!} \\ &=\frac 13 \left(\sum_{k=0}^\infty \frac{3^k e^{-3}}{k!}-e^{-3}\right) \end{aligned}$

But the sum is exactly the sum of probabilities of a Poisson distribution with mean 3. So it equals 1.

Therefore,
$\mathbb{P}(Y\leq 1)=\frac 13 ~ (1-e^{-3})$

Looks good to you ?