Timeline for explicitly compute conditional expectation
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2019 at 13:47 | vote | accept | CommunityBot | ||
| May 25, 2018 at 9:10 | comment | added | Kavi Rama Murthy | Put $t=0$ in the equation you have derived to get $g(0)=\frac 1 3$. For $t>0$ differentiate the equation to get $g(t)=t^{2}$. Hence $g(X)=\frac 1 3$ if $X=0$ and $X^{2}$ if $X>0$. These two facts can be expressed in one equation as $g(X)=\frac I 3 I_{X=0}+X^{2}$. | |
| May 25, 2018 at 7:50 | comment | added | user562844 | I come to the following solution: $\frac{1}{3}t^{3} + \frac{1}{3} = g(0) + \int_{0}^{t}g(max[0,x])dx$. It Looks similar to your answer, but when I differentiate with respect to t, then I get $g(\max[0,t]) = t^{2}$. So $\frac{1}{3}$ is missing | |
| May 24, 2018 at 14:59 | comment | added | user562844 | okay...then I get $\int_{-1}^{t}x^{2}dx = \int_{-1}^{t}f(\max[0,x])dx$. But now I have no plan how to proceed | |
| May 24, 2018 at 9:59 | comment | added | Kavi Rama Murthy | @gregor the definition says $\int_A x^{2} \, dx = \int_A f(X) \, dx$ for all sets $A$ of the type $\{X \leq t\}$ i.e. of the type $\{\max \{0,x\} \leq t \}$, $t \geq 0$. Use this equation to find $f$. On the right side you will get integral over $(-1,t)$, not $(-\sqrt t , \sqrt t)$. | |
| May 24, 2018 at 9:27 | comment | added | user562844 | is my approach correct? | |
| May 24, 2018 at 5:34 | comment | added | Kavi Rama Murthy | @gregor The answer is $E(Y|X)=\frac 1 3 I_{X=0} +X^{2}$ | |
| May 23, 2018 at 20:38 | comment | added | user562844 | my Approach is the following: $\int_{1}^{t}x^{2}dx = \int_{-\sqrt{t}}^{\sqrt{t}}f(max[0,x])dx$. But now I have no idea how to proceed | |
| May 23, 2018 at 20:25 | comment | added | user562844 | can you please also compute $\mathbb{E}(Y|X)$ | |
| May 18, 2018 at 15:43 | comment | added | user562844 | Thank you! I have difficulty with the theory of conditional expectation. Have you a good reference where I can learn from? Something with intuition behind it | |
| May 18, 2018 at 9:11 | history | answered | Kavi Rama Murthy | CC BY-SA 4.0 |