Expression for variance of compound Poisson process was included
L.V.Rao
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In the definition of the compound Poisson process, there is an underlying assumption that, as hinted by @Xian, the random variables $$X_{i}$$'s are i.i.d. as well as independent of $$N(t)$$. To compute $$E[S(t)]$$, consider $$\begin{eqnarray*} % \nonumber to remove numbering (before each equation) E\left[S(t)|N(t)=n\right] &=& E\left[X_{1}+X_{2}+\cdots +X_{N(t)}|N(t)=n\right] \\ &=& E\left[X_{1}+X_{2}+\cdots +X_{n}\right]\\ & & \quad\quad\quad\quad\quad\mbox{ since X's and N(t) are independent RV's}\\ &=& n E[X], \quad\quad\mbox{ since X_{i}'s are iid RV's} \end{eqnarray*}$$ Multiplying on both sides by $$P\left\{N(t)=n\right\}$$ and taking summation over all possible values of $$n$$, we get, $$\begin{eqnarray*} E\left[S(t)\right] &=& \sum_{n}\underbrace{E\left(S(t)|N(t)=n\right)}_{n\cdot E[X]} P\left\{N(t)=n\right\}\\ &=& \sum_{n} n\cdot E[X]\cdot P\left\{N=n\right\}\\ &=&E(X)\cdot \sum_{n}n\cdot P\left\{N=n\right\}\\ E\left[S(t)\right] &=& E[X]\cdot E[N(t)] = (\lambda t) \cdot E[X] \end{eqnarray*}$$ since $$\{N(t), t\geq 0\}$$ is Poisson process, $$E[N(t)] = \lambda t$$.

In order to find an expression for $$Var[S(t)]$$, first find the second conditional moment of $$S(t)$$. $$\begin{eqnarray*} E(S(t)^{2}|N(t)=n) &=& E[(X_{1}+X_{2}+\cdots +X_{N(t)})^{2}|N(t)=n] \\ &=& E[(X_{1}+X_{2}+\cdots +X_{n})^{2}]\\ &=& E\left\{\sum_{i=1}^{n}X_{i}^{2}+2\sum_{i Now $$\begin{eqnarray*} Var(S(t)) &=& E(S(t)^{2}) - \left[E(S(t))\right]^{2} \\ &=& \sum_{n}E(S(t)^{2}|N(t)=n)P\{N(t)=n\}-[E(X)E(N(t))]^{2}\\ &=&\sum_{n}\left[nVar\left(X\right)+n^{2}\left(E\left(X\right)\right)^{2}\right] P\left\{N(t)=n\right\} -\left[E(X)E(N(t))\right]^{2}\\ &=& Var(X) \sum_{n}nP\left\{N(t)=n\right\} + \left(E\left(X\right)\right)^{2}\sum_{n}n^{2}P\left\{N(t)=n\right\} - \left[E(X)E(N(t))\right]^{2}\\ &=&E(N(t))Var(X) + \left(E\left(X\right)\right)^{2} \left[E(N(t)^{2})-(E(N(t)))^{2}\right]\\ & = & E[N(t)\cdot Var[X] + Var[N(t)]\cdot (E[X])^{2}\nonumber\\ & =& \lambda t\cdot Var[X] + \lambda t\cdot (E[X])^{2}\nonumber\\ &=& \lambda t\cdot Var[X] + (E[X])^{2}]\nonumber\\ Var[S(t)]&=& \lambda t\cdot E[X^{2}] \end{eqnarray*}$$$$\begin{eqnarray*} Var(S(t)) &=& E(S(t)^{2}) - \left[E(S(t))\right]^{2} \\ &=& \sum_{n}E(S(t)^{2}|N(t)=n)P\{N(t)=n\}-[E(X)E(N(t))]^{2}\\ &=&\sum_{n}\left[nVar\left(X\right)+n^{2}\left(E\left(X\right)\right)^{2}\right] P\left\{N(t)=n\right\} -\left[E(X)E(N(t))\right]^{2}\\ &=& Var(X) \sum_{n}nP\left\{N(t)=n\right\} + \left(E\left(X\right)\right)^{2}\sum_{n}n^{2}P\left\{N(t)=n\right\} - \left[E(X)E(N(t))\right]^{2}\\ &=&E(N(t))Var(X) + \left(E\left(X\right)\right)^{2} \left[E(N(t)^{2})-(E(N(t)))^{2}\right]\\ & = & E[N(t)\cdot Var[X] + Var[N(t)]\cdot (E[X])^{2}\nonumber\\ & =& \lambda t\cdot Var[X] + \lambda t\cdot (E[X])^{2}\nonumber\\ &=& \lambda t\cdot [Var[X] + (E[X])^{2}]\nonumber\\ Var[S(t)]&=& \lambda t\cdot E[X^{2}] \end{eqnarray*}$$

Expression for variance of compound Poisson process was included
L.V.Rao
• 1.8k
• 13
• 22

In the definition of the compound Poisson process, there is an underlying assumption that, as hinted by @Xian, the random variables $$X_{i}$$'s are i.i.d. as well as independent of $$N(t)$$. To compute $$E[S(t)]$$, consider $$\begin{eqnarray*} % \nonumber to remove numbering (before each equation) E\left[S(t)|N(t)=n\right] &=& E\left[X_{1}+X_{2}+\cdots +X_{N(t)}|N(t)=n\right] \\ &=& E\left[X_{1}+X_{2}+\cdots +X_{n}\right]\\ & & \quad\quad\quad\quad\quad\mbox{ since X's and N(t) are independent RV's}\\ &=& n E[X], \quad\quad\mbox{ since X_{i}'s are iid RV's} \end{eqnarray*}$$ Multiplying on both sides by $$P\left\{N(t)=n\right\}$$ and taking summation over all possible values of $$n$$, we get, $$\begin{eqnarray*} E\left[S(t)\right] &=& \sum_{n}\underbrace{E\left(S(t)|N(t)=n\right)}_{n\cdot E[X]} P\left\{N(t)=n\right\}\\ &=& \sum_{n} n\cdot E[X]\cdot P\left\{N=n\right\}\\ &=&E(X)\cdot \sum_{n}n\cdot P\left\{N=n\right\}\\ E\left[S(t)\right] &=& E[X]\cdot E[N(t)] = (\lambda t) \cdot E[X] \end{eqnarray*}$$ since $$\{N(t), t\geq 0\}$$ is Poisson process, $$E[N(t)] = \lambda t$$.

In order to find an expression for $$Var[S(t)]$$, first find the second conditional moment of $$S(t)$$. $$\begin{eqnarray*} E(S(t)^{2}|N(t)=n) &=& E[(X_{1}+X_{2}+\cdots +X_{N(t)})^{2}|N(t)=n] \\ &=& E[(X_{1}+X_{2}+\cdots +X_{n})^{2}]\\ &=& E\left\{\sum_{i=1}^{n}X_{i}^{2}+2\sum_{i Now $$\begin{eqnarray*} Var(S(t)) &=& E(S(t)^{2}) - \left[E(S(t))\right]^{2} \\ &=& \sum_{n}E(S(t)^{2}|N(t)=n)P\{N(t)=n\}-[E(X)E(N(t))]^{2}\\ &=&\sum_{n}\left[nVar\left(X\right)+n^{2}\left(E\left(X\right)\right)^{2}\right] P\left\{N(t)=n\right\} -\left[E(X)E(N(t))\right]^{2}\\ &=& Var(X) \sum_{n}nP\left\{N(t)=n\right\} + \left(E\left(X\right)\right)^{2}\sum_{n}n^{2}P\left\{N(t)=n\right\} - \left[E(X)E(N(t))\right]^{2}\\ &=&E(N(t))Var(X) + \left(E\left(X\right)\right)^{2} \left[E(N(t)^{2})-(E(N(t)))^{2}\right]\\ & = & E[N(t)\cdot Var[X] + Var[N(t)]\cdot (E[X])^{2}\nonumber\\ & =& \lambda t\cdot Var[X] + \lambda t\cdot (E[X])^{2}\nonumber\\ &=& \lambda t\cdot Var[X] + (E[X])^{2}]\nonumber\\ Var[S(t)]&=& \lambda t\cdot E[X^{2}] \end{eqnarray*}$$

hope, this new version will be more clear.
L.V.Rao
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If $$\{N(t),t\geq 0\}$$ is aIn the definition of the compound Poisson Process with intensity parameter $$\lambda$$process, then it probability functionthere is givenan underlying assumption that, as hinted by @Xian, the random variables $$p_{n}(t)=P\{N(t)=n\}=e^{-\lambda t}\dfrac{(\lambda t)^{n}}{n!},\mbox{ for }n=0,1,2,\cdots$$ Its mean value function$$X_{i}$$'s are $$E\left\{N(t)\right\}=\lambda t$$i.i.d. as well as independent of $$N(t)$$.

If the process To compute $$\{S(t)=\sum_{i=1}^{N(t)}X_{i},\;t\geq 0\}$$$$E[S(t)]$$, consider $$\begin{eqnarray*} % \nonumber to remove numbering (before each equation) E\left[S(t)|N(t)=n\right] &=& E\left[X_{1}+X_{2}+\cdots +X_{N(t)}|N(t)=n\right] \\ &=& E\left[X_{1}+X_{2}+\cdots +X_{n}\right]\\ & & \quad\quad\quad\quad\quad\mbox{ since X's and N(t) are independent RV's}\\ &=& n E[X], \quad\quad\mbox{ since X_{i}'s are iid RV's} \end{eqnarray*}$$ Multiplying on both sides by $$P\left\{N(t)=n\right\}$$ and taking summation over all possible values of $$n$$, we get, $$\begin{eqnarray*} E\left[S(t)\right] &=& \sum_{n}\underbrace{E\left(S(t)|N(t)=n\right)}_{n\cdot E[X]} P\left\{N(t)=n\right\}\\ &=& \sum_{n} n\cdot E[X]\cdot P\left\{N=n\right\}\\ &=&E(X)\cdot \sum_{n}n\cdot P\left\{N=n\right\}\\ E\left[S(t)\right] &=& E[X]\cdot E[N(t)] = (\lambda t) \cdot E[X] \end{eqnarray*}$$ since $$\{N(t), t\geq 0\}$$ is a compound Poisson Processprocess, then

$$E\left\{S(t)\right\}=E\left\{N(t)\right\}\cdot E\left\{X\right\}, since \;\;X_{i}'s \mbox{ are iid rv.s}$$ $$E\left\{S(t)\right\}=(\lambda t)E(X)$$$$E[N(t)] = \lambda t$$.

L.V.Rao
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