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 January 24th, 2018, 01:00 AM #1 Newbie   Joined: Jan 2018 From: ChengDu China Posts: 4 Thanks: 0 a strange joint probability problem For example, F (x, y) is a two-dimensional joint probability distribution function, where X is a normal normal distribution, Y is 0-1 binomial distribution with parameter q. How to find the expectation of F (x, y)? thanks for solving this problem Last edited by fun; January 24th, 2018 at 01:17 AM. Reason: make the question more clearly
 January 24th, 2018, 09:38 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1220 your question doesn't make sense as written. $F(x,y)$ is not a random variable and thus it doesn't have an expectation. you can find $E[X]$, or $E[Y]$, or $E[g(X,Y)]$ as $X$ and $Y$ are random variables So what exactly are you trying to find? Thanks from fun
January 24th, 2018, 05:54 PM   #3
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 Originally Posted by romsek your question doesn't make sense as written. $F(x,y)$ is not a random variable and thus it doesn't have an expectation. you can find $E[X]$, or $E[Y]$, or $E[g(X,Y)]$ as $X$ and $Y$ are random variables So what exactly are you trying to find?
thanks!
that's my fault, I will make the question clearer！
suppose
X,Y are both random variables
F(X,Y) is a function on X Y
X obeys the normal distribution with mean m, standard deviation n
Y obeys poisson distribution with parameter p
so how to caculate E[F(X,Y)]

January 24th, 2018, 08:12 PM   #4
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 Originally Posted by fun thanks! that's my fault, I will make the question clearer！ suppose X,Y are both random variables F(X,Y) is a function on X Y X obeys the normal distribution with mean m, standard deviation n Y obeys poisson distribution with parameter p so how to caculate E[F(X,Y)]
We have to assume that $X$ and $Y$ are independent as you've specified just their marginals.

let $Z=F(X,Y)$

$\large E[Z] = \displaystyle \sum \limits_{y=0}^\infty~\int_{-\infty}^{\infty}~F(x,y) \dfrac{1}{\sqrt{2\pi}n}e^{-\frac{(x-m)^2}{2n^2}}\cdot \dfrac{p^y e^{-p}}{y!}~dx$

This might simplify a little bit without specifying $F(X,Y)$ but not much.

January 25th, 2018, 12:13 AM   #5
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 Originally Posted by romsek We have to assume that $X$ and $Y$ are independent as you've specified just their marginals. let $Z=F(X,Y)$ $\large E[Z] = \displaystyle \sum \limits_{y=0}^\infty~\int_{-\infty}^{\infty}~F(x,y) \dfrac{1}{\sqrt{2\pi}n}e^{-\frac{(x-m)^2}{2n^2}}\cdot \dfrac{p^y e^{-p}}{y!}~dx$ This might simplify a little bit without specifying $F(X,Y)$ but not much.
thanks again!
you mean that
to solve this kind of question
we don't need use Lebesgue integral，Riemann integral is enough?

January 25th, 2018, 08:32 AM   #6
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 Originally Posted by fun thanks again! you mean that to solve this kind of question we don't need use Lebesgue integral，Riemann integral is enough?
obviously it's going to depend on $F(x,y)$

I suppose you could come up with one that would require a Lebesgue integral but in general you won't need one. Each value of $Y$ provides, depending on $F(x,y)$ an Riemann integrable function in $x$

January 25th, 2018, 05:54 PM   #7
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 Originally Posted by romsek obviously it's going to depend on $F(x,y)$ I suppose you could come up with one that would require a Lebesgue integral but in general you won't need one. Each value of $Y$ provides, depending on $F(x,y)$ an Riemann integrable function in $x$
thanks a lot!
can you give me an example that we need to use Lebsgue intergral?
Let$F[X,Y]=tan(X+Y)$
can we use Riemann intergral?

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