Timeline for Expected Value of a Function of Random Variable
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2024 at 17:07 | comment | added | John Dawkins | You can also express the Law of Total Probability (in the context of breaking things down using 3 disjoint events $A,B,C$ with union all of the sample space) as $$ E[Y] = E[Y; A]+E[Y; B]+E[Y; C]. $$ It may or may not be helpful in a given situation to further write $E[Y; B]$ as $E[Y\mid B]\cdot P[B]$. This is helpful, for example, in the second of your three terms. | |
| Oct 2, 2024 at 18:32 | comment | added | Scentless Apprentice | I'm wondering if the answer is because each conditional pdf is supposed to be weighted by the probability of the conditioning event. So the reason my initial solution worked out is because $$E[X-1|1\leq X \leq 6]\cdot P(1\leq X \leq 6) ={\int_{1}^{6} \frac{f_{X}(x)}{P(1\leq X \leq 6)} dx}\cdot P(1 \leq X \leq 6) = \int_{1}^{6} f_{X}(x) dx$$ | |
| Oct 2, 2024 at 18:05 | comment | added | Scentless Apprentice | I'm sorry, now I'm totally confused. I initially found $E[Y]$ by summing the expectations of the three events that define $Y$ and obtaining $2(e^{-.5} - e^{-3})$ as the solution. However, the theorem of total expectation says that each event should be weighted by its respective probability, which I didn't do. So instead of $$E[Y] = E[0 | 0\leq X \leq 1] + E[X-1 | 1 < X \leq 6] + E[5 | X>6]$$ it should be $$E[Y] = E[0 | 0\leq X \leq 1]\cdot P(0 \leq X \leq 1) + E[X-1 | 1< X \leq 6] \cdot P(1 < X \leq 6)+ E[5 | X>6] \cdot P(X>6)$$ A completely different answer! Where am I going wrong? | |
| Oct 2, 2024 at 15:19 | comment | added | Scentless Apprentice | Oh, I think I see. The work you've shown rewrites the expectation for the event {1 < X <=6} in a way that allows you to exploit the memorylessness property of the distribution. Is that correct? | |
| Oct 2, 2024 at 14:56 | comment | added | John Dawkins | The three terms correspond to the three non-overlapping events $\{X\le 1\}$, $\{1<X\le 6\}$, and $\{X>6\}$. | |
| Oct 2, 2024 at 14:50 | comment | added | Scentless Apprentice | Hi John -- Thanks for the answer! If I'm not mistaken, this "technique" is just an application of the total expectation theorem, stated in my text as E[X] = \sum_{i=1}^{n} P(A_{i})E[X|A_{i}], with A_{i}'s disjoint. But in the second and third terms above, it seems they overlap. I understand the motivation to use P(X>1) in combination with the function X-1, but doesn't that overlap with P(X>6)? I appreciate your help and patience! | |
| Sep 27, 2024 at 23:46 | vote | accept | Scentless Apprentice | ||
| Sep 27, 2024 at 23:07 | history | edited | John Dawkins | CC BY-SA 4.0 | added 5 characters in body |
| Sep 27, 2024 at 22:54 | history | answered | John Dawkins | CC BY-SA 4.0 |