New answers tagged

-1

$$E\left[ \sum_{n=1}^{\infty} 1_{|X| > n} \right]$$ $$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{|X| > n} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\lfloor|X|\rfloor > n} \right]$$ $$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{\Omega} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\emptyset} \right]$$ $$ = E\left[ \sum_{n=1}^{\lfloor|X|\...


-1

You derailed on the third line.   Backtrack and try this: $$\begin{align} \varphi_{S_N}(t) = &~ \mathsf E(\mathscr e^{\imath t S_N}) \\[1ex]=&~ \mathsf E(\mathscr e^{\imath t \sum_{k=1}^N X_k}) \\[1ex]=&~ \mathsf E(\prod_{k=1}^N\mathscr e^{\imath t X_k}) & \because~e^{a+b+c+\ldots}=e^a\cdot e^b\cdot e^c\cdots \\[1ex]=&~ \mathsf E(\...


1

$\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}$The law of total expectation tells you that $$ \E(e^{itS_N}) = \E(\E(e^{itS_N}\mid N)). $$ So you have \begin{align} \varphi_{S_N}(t) = \E(e^{itS_N}) & = \E(\E(e^{itS_N}\mid N)) = \E\left( \E\left( e^{\sum_{j=1}^N it X_j } \mid N \right) \right) \\[10pt] & = \E\left( \E \left( \...


1

I think earlier in your calculation some things got confusing. You should just find: $$\phi_S(t)=E[e^{it \sum_{j=1}^N X_j}]=\sum_{n=0}^\infty E \left [e^{it \sum_{j=1}^N X_j} \mid N=n \right ] P[N=n] \\ = \sum_{n=0}^\infty E \left [ e^{it \sum_{j=1}^n X_j} \right ] P[N=n] \\ = \sum_{n=0}^\infty \phi_X(t)^n P[N=n]$$ using the total expectation formula and ...


0

Just a thought: By setting the inequality as an equality and substituting $y=0$, $(x-8)^2+0+14\sin(x-8)+14\cdot 0=30$ I found the two intersections of the surface at $y=0$ at around $x\approx 3.78962$ and $x\approx 13.9171$. I wonder if it's possible to calculate the volume by using the same approach that you used to compute the area. However, the ...



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