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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: 2,638 Thanks: 1475 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? Tags calculate, distribution, expectation, joint, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post IndependentThinker Advanced Statistics 1 February 27th, 2015 03:01 PM batman350z Advanced Statistics 3 April 23rd, 2012 03:50 PM xdeathcorex Probability and Statistics 0 April 22nd, 2011 08:55 AM meph1st0pheles Advanced Statistics 1 March 23rd, 2010 06:47 PM adv Advanced Statistics 1 April 9th, 2009 10:22 AM

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