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January 24th, 2018, 12:00 AM   #1
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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 12:17 AM. Reason: make the question more clearly
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January 24th, 2018, 08:38 AM   #2
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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?
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January 24th, 2018, 04:54 PM   #3
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Quote:
Originally Posted by romsek View Post
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)]
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January 24th, 2018, 07:12 PM   #4
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Quote:
Originally Posted by fun View Post
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.
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January 24th, 2018, 11:13 PM   #5
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Quote:
Originally Posted by romsek View Post
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?
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January 25th, 2018, 07:32 AM   #6
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Quote:
Originally Posted by fun View Post
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 $x$
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January 25th, 2018, 04:54 PM   #7
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Quote:
Originally Posted by romsek View Post
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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