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
Hello,
$\displaystyle \mathbb{P}(Y\leq 1)=\sum_{k=0}^\infty \mathbb{P}(Y\leq 1 \mid X=k)\mathbb{P}(X=k)=\sum_{k=0}^\infty \mathbb{P}(Y\leq 1 \mid X=k) \cdot\frac{3^k e^{-3}}{k!}$
But then, which distribution follows the random variable $\displaystyle Y\mid X=k$ ? It's a uniform distribution over $\displaystyle [0,k+1]$
Where the pdf is $\displaystyle f(y)=\begin{cases}\frac{1}{k+1} & \text{ if } y\in[0,k+1] \\ 0 & \text{ otherwise} \end{cases}=\frac{1}{k+1} \cdot \bold{1}_{[0,k+1]}(y)$
$\displaystyle \mathbb{P}(Y\leq 1\mid X=k)=\int_0^1 \frac{1}{k+1} ~dy$ (because $\displaystyle k+1\geq 1$ almost surely, we get these boundaries)
Which is $\displaystyle \mathbb{P}(Y\leq 1\mid X=k)=\frac{1}{k+1}$
Hence
$\displaystyle \begin{aligned}
\mathbb{P}(Y\leq 1)
&=\sum_{k=0}^\infty \frac{3^k e^{-3}}{(k+1)!} \\
&=\frac 13 \sum_{k=0}^\infty \frac{3^{k+1}e^{-3}}{(k+1)!} \\
&=\frac 13 \sum_{k=1}^\infty \frac{3^k e^{-3}}{k!} \\
&=\frac 13 \left(\sum_{k=0}^\infty \frac{3^k e^{-3}}{k!}-e^{-3}\right)
\end{aligned}$
But the sum is exactly the sum of probabilities of a Poisson distribution with mean 3. So it equals 1.
Therefore,
$\displaystyle \mathbb{P}(Y\leq 1)=\frac 13 ~ (1-e^{-3})$
Looks good to you ?