Timeline for Suppose X and Y are independent Poisson random variables with respective parameters $\lambda$ and 2$\lambda$. Find $E[Y-2*X|X+Y=10]$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2020 at 17:23 | answer | added | Adam Little | timeline score: 1 | |
| Feb 7, 2020 at 23:44 | comment | added | whuber♦ | The point where the conditional probability was replaced by a probability is invalid: it's tantamount to supposing the condition $X+Y=10$ doesn't affect anything. One way to see why not is to replace "10" everywhere by "0". You would wind up concluding $E[Y-2X\mid X+Y=0]=0-3\lambda$ but that's obviously incorrect because $X+Y=0$ implies $X=Y=0,$ whence $Y-2X=0$ whose expectation is $0,$ not $-3\lambda.$ | |
| Feb 7, 2020 at 23:23 | comment | added | That One Dude Mike | Cool - I've also had someone suggest E[Y-2X|X+Y=10]=E[Y-2X|Y=10-X]=E[10-X-2X]=10-3*E[X]=10-3*lambda (since E[X]=lambda). Would you know if those statements actually equivalent, or am I going down the wrong path? | |
| Feb 7, 2020 at 21:56 | comment | added | whuber♦ | The information you need (explained in several different ways) is at stats.stackexchange.com/questions/429564/…. Although your answer is correct, the notation makes no sense at several points, which is what your professor may be objecting to. In particular, most people would interpret expressions like "$\operatorname{Bin}(10,2/3)$" as referring either to distributions or random variables, but not to expectations. | |
| Feb 7, 2020 at 21:02 | history | asked | That One Dude Mike | CC BY-SA 4.0 |