Solved – Characteristic Function of a Compound Poisson Process

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

Let $lambda>0$ and $Nsimtext{Poisson}(lambda T)$. Also, ${X_i}_{i=1}^N$ are i.i.d. and independent of $N$. And ${U_i}_{i=1}^N$ are i.i.d., $U_isimtext{Uniform}([0,T])$, and independent from $X_i,N$. Define:
Y_tequivsum_{i=1}^Nmathbb{1}_{{U_ileq t}}X_i, 0leq tleq 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:

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:

mathbb{E}(e^{iuY_1})&=sum_nP(N=n)mathbb{E}(e^{iuY_1}mid N=n)\
&=sum_nP(N=n)prod_{j=1}^nmathbb{E}(e^{iumathbb{1}_{{U_jleq 1}}X_j})quadtext{(by independence)}\
&=sum_n P(N=n)prod_{j=1}int e^{iux}f(x)dxquadtext{(by uniform)}

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

I was missing the knowledge of the exponential series: $$ e^x=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots $$ I also made a mistake when I separated the uniform in the expectation. Fixing these problems: $$ begin{align} mathbb{E}(e^{iuY_1})&=sum_nP(N=n)mathbb{E}(e^{iuY_1}mid N=n)\ &=sum_nP(N=n)prod_{j=1}^nmathbb{E}(e^{iumathbb{1}_{{U_jleq 1}}X_j})quadtext{(by independence)}\ &=sum_n P(N=n)left(mathbb{E}(e^{iumathbb{1}_{{U_1leq 1}}X_1})right)^nquadtext{(by i.i.d.)}\ &=sum_{n=0}^inftyfrac{(lambda T)^n e^{-(lambda T)}}{n!}left(mathbb{E}(e^{iumathbb{1}_{{U_1leq 1}}X_1})right)^nquadtext{(by Poisson)}\ &=e^{-(lambda T)}cdot e^{(lambda T)mathbb{E}(e^{iumathbb{1}_{{U_1leq 1}}X_1})}quadtext{(by the exponential series)} end{align} $$ We can calculate the expectation by conditioning on the uniform: $$ mathbb{E}(e^{iumathbb{1}_{{U_1leq 1}}X_1})=frac{T-1}{T}+frac{1}{T}int e^{iux}f(x)dx $$ Substituting and doing some algebra we get the answer: $$ begin{align} mathbb{E}(e^{iuY_1})&=e^{lambda int (e^{iux}-1)f(x)dx} end{align} $$

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