Timeline for Prove a result on expectation with 2 random variables
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2020 at 17:38 | vote | accept | Arshdeep | ||
| Jul 4, 2020 at 17:13 | comment | added | doubled | @ArshdeepSinghDuggal no I meant that, but added a * sign to make clear what we are doing. I left $S$ as is... depending on $S$, you could write $S(T_1)$ and $S(T_2)$ but you can't remove the indicator function. | |
| Jul 4, 2020 at 17:11 | history | edited | doubled | CC BY-SA 4.0 |
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| Jul 3, 2020 at 22:24 | comment | added | Arshdeep | Did you mean to write $E[S]=E[S1[T=1]]+E[S2[T=2]]$? | |
| Jul 3, 2020 at 22:16 | history | edited | doubled | CC BY-SA 4.0 |
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| Jul 3, 2020 at 22:14 | comment | added | doubled | I made sure to state that I'm taking $Y$ to be discrete but will further emphasize at start of my post! The second comment of the post linked in the above comment explains it nicely: "If {Y=y} doesn't have measure zero and can therefore be re-scaled into a probability space, you're just talking about the restriction of X to that space, which is a random variable whose density is the conditional density." | |
| Jul 3, 2020 at 22:11 | comment | added | gunes | Given $Y=y$, $X$ has a density, but this doesn't mean that we can define a RV $Z=X|Y=y$ Similar confusions are discussed in the following posts: math.stackexchange.com/questions/612468/… math.stackexchange.com/questions/711890/… | |
| Jul 3, 2020 at 22:05 | history | edited | doubled | CC BY-SA 4.0 |
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| Jul 3, 2020 at 21:59 | history | edited | doubled | CC BY-SA 4.0 |
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| Jul 3, 2020 at 21:47 | history | answered | doubled | CC BY-SA 4.0 |