Let $N(t)$ be a standard Poisson process with intensity $\lambda>0$. As we know, for $n \geq 1$ we can define the time of $n$th event as $ S_n = \sum_{k=1}^n T_k, \; S_0 = 0, $ where $(T_k)$ are iid exponentially distributed random variables with mean $1/\lambda$. We can rewrite the definition of our Poisson process in terms of $(S_k)$, as $ N(t) = \sum_{n=1}^\infty \mathbb{1}_{(0,t]}(S_n). $
Now, let $(D_k)$ be iid exponentially distributed random variables with mean $1/\mu$, independent of $(T_k)$. We define $$ \tilde{S_n} = \sum_{k=1}^n T_k + D_n. $$ So basically, we randomly translate time of every event of our Poisson process. Now let $\tilde{N}(t)$ be a process that counts events $\tilde{S}_n$ that occurred on the interval $(0,t]$, that is $ \tilde{N}(t) = \sum_{n=1}^\infty \mathbb{1}_{(0,t]}(\tilde{S}_n). $
As far as I know, it is not Poisson process anymore. But can we say something about the distribution of $\tilde{N}(t)$? The major problem I stumble upon is that the inequality $\tilde{S}_n \leq \tilde{S}_{n+1}$ does not hold a.s., as in classical Poisson process. I have also tried conditioning: $$ \mathbb{P}(\tilde{N}(t) = n) = \sum_{k=0}^n \mathbb{P}(\tilde{N}(t)=n, N(t) = k) \mathbb{P}(N(t)=k), $$ since $N(t) \leq \tilde{N}(t)$ a.s. The event $\mathbb{P}(\tilde{N}(t)=n, N(t) = k)$ means that there are $n-k$ events on the interval $(0,t]$ that has been translated to $(t, \infty)$. However, I don't know how to deal with it. Memoryless property of exponential distribution may seem to be helpful in this problem.