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the expectation of products is a product of expectations, given independence
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The definition of a compound Poisson process and its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\sum_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\\ &=\sum_n P(N=n)\sum_{j=1}\int e^{iux}f(x)dx \end{align} $$$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\prod_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\quad\text{(by independence)}\\ &=\sum_n P(N=n)\prod_{j=1}\int e^{iux}f(x)dx\quad\text{(by uniform)} \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

The definition of a compound Poisson process and its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\sum_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\\ &=\sum_n P(N=n)\sum_{j=1}\int e^{iux}f(x)dx \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

The definition of a compound Poisson process and its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\prod_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\quad\text{(by independence)}\\ &=\sum_n P(N=n)\prod_{j=1}\int e^{iux}f(x)dx\quad\text{(by uniform)} \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

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The definition of thea compound Poisson process theand its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\sum_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\\ &=\sum_n P(N=n)\sum_{j=1}\int e^{iux}f(x)dx \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

The definition of the compound Poisson process the characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\sum_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\\ &=\sum_n P(N=n)\sum_{j=1}\int e^{iux}f(x)dx \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

The definition of a compound Poisson process and its characteristic function I have are the following:

Let $\lambda>0$ and $N\sim\text{Poisson}(\lambda T)$. Also, $\{X_i\}_{i=1}^N$ are i.i.d. and independent of $N$. And $\{U_i\}_{i=1}^N$ are i.i.d., $U_i\sim\text{Uniform}([0,T])$, and independent from $X_i,N$. Define: $$ Y_t\equiv\sum_{i=1}^N\mathbb{1}_{\{U_i\leq t\}}X_i, 0\leq t\leq T $$ Then $Y_t$ is a compound Poisson process with intensity parameter $\lambda$ and jump pdf $f(x)$.

The characteristic function of $Y_1$ is: $$ \mathbb{E}(e^{iuY_1})=e^{\lambda\int(e^{ix}-1)f(x)dx} $$

Note that the characteristic function I quoted above is for $Y_1$, not $Y_t$. I am trying to show the equality above. I currently have:

$$ \begin{align} \mathbb{E}(e^{iuY_1})&=\sum_nP(N=n)\mathbb{E}(e^{iuY_1}\mid N=n)\\ &=\sum_nP(N=n)\sum_{j=1}^n\mathbb{E}(e^{iu\mathbb{1}_{\{U_j\leq 1\}}X_j})\\ &=\sum_n P(N=n)\sum_{j=1}\int e^{iux}f(x)dx \end{align} $$

I am not sure how to proceed. Any tips? Thanks for helping! :D

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