Timeline for E[ | X - Y | ] where X and Y are independent Poisson random variable
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2021 at 9:31 | comment | added | Clement C. | Related: mathoverflow.net/questions/396126/… | |
| Dec 10, 2017 at 12:05 | review | Close votes | |||
| Dec 10, 2017 at 17:10 | |||||
| S Dec 10, 2017 at 10:45 | history | suggested | Martin Thoma | CC BY-SA 3.0 |
improve formatting
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| Dec 10, 2017 at 10:13 | review | Suggested edits | |||
| S Dec 10, 2017 at 10:45 | |||||
| Aug 5, 2012 at 13:33 | comment | added | Brendan McKay | The difference of two Poisson distributions is called a Skellam distribution and can be expressed in terms of modified Bessel functions. I don't see why the expectation of the absolute value should have a simple expression. | |
| Aug 5, 2012 at 11:35 | comment | added | Douglas Zare | In general, you can't express the expected absolute value that way. Consider $f(x) = x^5+2x-1$ on $[0,1]$. The average value of $f$ is easy to compute, but to compute the average value of $|f|$ you need to be able to express the root of $f$ on that interval. That can be an example of an absolute value of the difference between random variables where one is the constant $0$. | |
| Aug 5, 2012 at 11:16 | comment | added | mathsguy1 | No its not a homework question (I left school many years ago!). Its something I came across while working through some derivations in a Mathematical Finance book recently. Maybe I was on the wrong track, but it just seems E[|X - Y|] should be expressible in terms of E[X] and E[Y]. | |
| Aug 5, 2012 at 11:02 | comment | added | Douglas Zare | I agree that you should give more context. It's not clear whether to expect there to be a simple solution or not. The "variance or covariance term" suggestion seems wrong. Integrating or summing the absolute values of simple functions can produce much more complicated results. Anyway, for equal parameters, Mathematica sums over the region where $X\gt Y$ (half of the answer) and gets $\exp(-2 \lambda)(BesselI[1,2\lambda] + \lambda BesselI[2,2 \lambda])$. | |
| Aug 5, 2012 at 10:50 | comment | added | David Roberts♦ | In what context did this question arise? If homework or similar exercise, please see the FAQ for why this is not an appropriate forum for this question. Otherwise, I'm still not sure this is the site you are looking for. | |
| Aug 5, 2012 at 10:40 | history | asked | mathsguy1 | CC BY-SA 3.0 |