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I apologize if this question is not placed in the right place. But I am having a hard time to figure it out. It would be greatly appreciated if some one could help me out.

Assume that there are two processes $X_1(t),X_2(t)$ . The process $X_1(t)$ could take three possible values $x_{1,1}(t),x_{1,2}(t),x_{1,3}(t)$. Likewise, $X_2(t)$ could take three possible distinct values $x_{2,1}(t),x_{2,2}(t),x_{2,3}(t)$. Hence $(X_1(t),X_2(t))$ could take nine possible values $(x_{1,i}(t),x_{2,j}(t))$ for $i,j=1,2,3$. Assume that $P((X_1(t),X_2(t))=(x_{1,i}(t),x_{2,j}(t)))=p_{i,j}(t)$ which is known for $i,j=1,2,3$ Now let two correlated Poisson processes $N_i(t)$'s with intensity rate $\lambda_i t>0$

(more specifically $p(N_i(t)=k)=e^{-\lambda_it}\frac{(\lambda_it)^k}{k!}$)

Define two new processes: $X'_1(t):=X_1(t)+\sum_{i=1}^{N_1(t)}Y_i(t)$

and $X'_2(t)=X_2(t)+\sum_{i=1}^{N_2(t)}Y_j'(t)$

Where $\{Y_i(t)\}$ and $\{Y_i'(t)\}$ are sequences of i.i.d $N(\mu,\delta)$, and independent of $X_i$, $N_i$.

Note that there joint distribution of $(N_1,N_2)$ can be seen here https://math.stackexchange.com/questions/244989/correlated-poisson-distribution.

My question is : Can we find the probabilities (or at least approximate the probabilities) of $P((X'_1(t),X'_2(t))=(x_{1,i}(t),x_{2,j}(t))$ for $i,j=1,2,3$, and $P((X'_1(t),X'_2(t))\neq (x_{1,i}(t),x_{2,j}(t))$ based on the distribution of $(X_1,X_2)$ , $Y_i$, $Y_j'$, $N_1$ and $N_2$ ?

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To make notation a little bit lighter I am going to use $N=(N(t)_1,N(t)_2)$, $X=(X(t)_1,X(t)_2)$, $X'=(X'(t)_1,X'(t)_2)$. $x$ is any of the nine possible answers.

The only case when $Pr(X'=x) \neq 0$ is when $N=0$. If not $X'$ is continuous, because $Y$ is continuous, and $Pr(X'=x)=0$.

$ Pr(X'=x) = Pr(X'=x|N=0)Pr(N=0) + Pr(X'=x|N\neq 0)Pr(N\neq 0) = Pr(X'=x|N=0)Pr(N=0) + 0 = Pr(X=x|N=0)Pr(N=0) = Pr(X=x)Pr(N=0)$

Step 1: $Pr(A) = Pr(A \cap B) + Pr(A\cap B^{\complement})$

Step 2: we know that when $N\neq 0$ $Pr(X'=x|N \neq 0)=0$, as explained above.

Step 3: if $N=0$ then $X'=X$

Step 4: $N$ and $X$ are independent. $Pr(N=0)$ can be calculated using the link you provided.

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    $\begingroup$ Thanks Vincent. I forgot to say that $X$ AND $N$ are independent. $\endgroup$ – Nguyen Jul 4 '14 at 13:51

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