I am trying to find a formula to find the expected value of a RV that is mixed (continuous and discrete).
I defined the nonnegative RV $Y$ as a function of the absolutely continuous RV $X$. $Y=X$ if $X$ takes Values in $[a,b]$ and $Y=C$ else, while $C$ is a discrete RV.
if $\mathbb{I}$ denotes the indicator function i can write:
$Y=\mathbb{I}_{[a,b]}(X)X+\mathbb{I}_{[0,a)\cup(b,\infty)}(X)C$ (is this right?)
with the definition of the expected Value from measure theory i wrote (if it exists):
$\mathbb{E}[Y]=\mathbb{E}[g(X)]=\int_\mathbb{R}\mathbb{I}_{[a,b]}(X)X+\mathbb{I}_{[0,a)\cup(b,\infty)}(X)Cd\mathbb{P}=\int_\mathbb{R}\mathbb{I}_{[a,b]}(X)Xd\mathbb{P}+\int_\mathbb{R}\mathbb{I}_{[0,a)\cup(b,\infty)}(X)Cd\mathbb{P}$.
I think the first Integral can be solved like this: $X:\mathbb{R}\rightarrow\mathbb{R}$ is defined on the measure space $(\mathbb{R},\mathcal{A}_1,\mathbb{P})$.
define the pushforward measure $P_X(A):=\mathbb{P}(X^{-1}(A))$ and $Z:\mathbb{R}\rightarrow\mathbb{R}$ with $Z(x)=x$ on $(\mathbb{R},\mathcal{A}_2,P_X)$.
By using the change of variable theorem:
$\int_\mathbb{R}\mathbb{I}_{[a,b]}(X)Xd\mathbb{P}=\int_{[a,b]}Z(X(x))dP_X(x)=\int_{[a,b]}xdP_X=\int_{[a,b]}xf_X(x)dx$
when $f_X$ denotes the density of $X$.
I had trouble finding the second integral: $\int_\mathbb{R}\mathbb{I}_{[a,0)\cup(b,\infty)}(X)Cd\mathbb{P}$.
I know that i can write $C$ as a simple function, what makes it easy to integrate it. But i dont know how to deal with the indicator function.
I think the Integral should be something like:
$\mathbb{P}(X\in[0,a)\cup(b,\infty))\sum_{i\in\mathbb{N}}c_i\mathbb{P}(C=c_i)$ when $c_i$ is the realisation of $C$.
I would appreciate any help or tips. Im sorry if anything ist unclear formulated since english is not my First language and i am very new to measure theory.
You have to use joint pdf $P(C=c,X=x) = P(C=c) P(X=x)$.
$$E(g(X,C)) = \int \sum_c \ g(x,c) \times P(C=c,X=x) \ \ dx$$ $$E(1_S(X) C) = \int \sum_c \ 1_S(x) \times c \times P(C=c,X=x) \ \ dx$$ $$E(1_S(X) C) = \int \sum_c 1_S(x) \times c \times P(C=c) P(X=x)\ \ dx$$ $$E(1_S(X) C) = \sum_c c P(C=c) \int 1_S(x) P(X=x) \ \ dx $$ $$E(1_S(X) C) = P(X \in S) \times \sum_c c P(C=c)$$ $$E(1_S(X) C) = P(X \in S) \times E(C)$$