Let $Y\sim \operatorname{Geo}(p)$ with $P(Y=k)=p(1-p)^k$. Furthermore, $\hat{Y}\sim \operatorname{NegBin}(2,p)$, i. e. $P(\hat{Y}=k)=kp^2(1-p)^{k-1}$.
I want to find $P(\lfloor U\widehat{Y} \rfloor=k)$ where $U\sim \mathcal{U}([0,1])$ and independent of $\widehat{Y}$.
My attempt:
$P(\lfloor U\widehat{Y} \rfloor)=P(k\leq U\widehat{Y}<k+1)$ Is it true that $$P(k\leq U\widehat{Y}<k+1)=P(k\leq U<k+1)\cdot P(k\leq \widehat{Y} <k+1)?$$
If so, the first term would be $=1$ but the second not equal to $p(1-p)^k$. But I have to prove that $\lfloor U\widehat{Y}\rfloor$ has the same distribution as $Y$.
No, $P[k \leq U \hat{Y} < k + 1] \neq P[k \leq U < k + 1] \cdot P[k \leq \hat{Y} < k + 1]$. Otherwise you would get a contradiction because for all integers $k \geq 1$, $P[k \leq U < k + 1] = 0$ since the support of $U$ is $[0, 1]$ so any event where $1 \leq U$ has probability $0$.
Instead use the law of total probability: for any fixed $k \in \mathbb N$ \begin{align*} P[k \leq U \hat{Y} < k + 1] &= \sum_{\hat{k} = 1}^\infty P[k \leq U \hat{Y} < k + 1 \ | \ \hat{Y} = \hat{k}] \cdot P[\hat{Y} = \hat{k}] \\ &= \sum_{\hat{k} = 1}^\infty P[k \leq U \hat{k} < k + 1 \ | \ \hat{Y} = \hat{k}] \cdot P[\hat{Y} = \hat{k}] \\ &= \sum_{\hat{k} = 1}^\infty P\Big[\frac{k}{\hat{k}} \leq U < \frac{k + 1}{\hat{k}} \ \Big| \ \hat{Y} = \hat{k}\Big] \cdot P[\hat{Y} = \hat{k}] \\ &= \sum_{\hat{k} = 1}^\infty P\Big[\frac{k}{\hat{k}} \leq U < \frac{k + 1}{\hat{k}}\Big] \cdot P[\hat{Y} = \hat{k}] \quad \text{ as } U, \hat{Y} \text{ independent} \end{align*} Try to work out the details of the sum above. To get started, notice that for integers $\hat{k} \leq k$, $ P\Big[\frac{k}{\hat{k}} \leq U < \frac{k + 1}{\hat{k}}\Big] = 0$ (again as $U$ cannot take values above $1$). So actually you can simplify the sum to go from $k + 1$ to $\infty$.