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Let $xi $ be a random variable in $(Omega, mathcal{F}, mathbb{P})$ distributed according to $mathrm{Unif}{-2,...,3}$. Hence,
$$mathbb{P}(xi = -2) = mathbb{P}(xi = -1) =mathbb{P}(xi = 0) =mathbb{P}(xi = 1) =mathbb{P}(xi = 2) =mathbb{P}(xi = 3) =1/6$$
Now let $mathcal{D}={D_1,D_2,D_3}$ where $D_1={xi<0}$ $D_2={xi in [0,2]}$ $D_3={ xi >2 }$.

I need to find:

1) $mathbb{E}(xi^2|mathcal{D})$ (The conditional expectation).

2) $mathbb{P}(xi<1|mathcal{D})$.

I am not sure, but I think that for 1) I should use formula $mathbb{E}(xi^2 |mathcal{D})(omega)= sum_{i} frac{mathbb{E}(xi^2 mathbf{1}_{D_i)}}{mathbb{P}(D_i)} mathbf{1}_{D_i} (omega) $. But how should I use it?

I assume that $mathbb E[xi^2midmathcal D]$ denotes the same as $mathbb E[xi^2midmathcalsigma(mathcal D)]$

$mathbb{E}left[xi^{2}midmathcal{D}right]$ is measurable wrt
$sigmaleft(mathcal{D}right)$ implying here the existence of coefficients
$a,b,c$ with $$mathbb{E}left[xi^{2}midmathcal{D}right]=amathbf{1}_{xi<0}+bmathbf{1}_{0leqxileq2}+cmathbf{1}_{xi>2}tag1$$

Secondly we have the equalities: $$mathbb{E}left[xi^{2}mathbf{1}_{D}right]=mathbb{E}left[mathbb{E}left[xi^{2}midmathcal{D}right]mathbf{1}_{D}right]text{ for every }Dinsigmaleft(mathcal{D}right)tag2$$

Substituting $left(1right)$ in $(2)$ we find for $D={xi<0}$, ${0leqxileq2}$ and ${xi>2}$:

• $mathbb{E}left[xi^{2}mathbf{1}_{xi<0}right]=aPleft(xi<0right)$




• $mathbb{E}left[xi^{2}mathbf{1}_{0leqxileq2}right]=bPleft(0leqxileq2right)$




• $mathbb{E}left[xi^{2}mathbf{1}_{xi>2}right]=cPleft(xi>2right)$

Or equivalently

• $a=mathbb{E}left[xi^{2}midxi<0right]=frac{1}{2}left(-2right)^{2}+frac{1}{2}left(-1right)^{2}=frac{5}{2}$


• $b=mathbb{E}left[xi^{2}mid0leqxileq2right]=frac{1}{3}0^{2}+frac{1}{3}1^{2}+frac{1}{3}2^{2}=frac{5}{3}$


• 
  $c=mathbb{E}left[xi^{2}midxi>2right]=3^{2}=9$.

So we end up with: $$mathbb{E}left[xi^{2}midmathcal{D}right]=frac{5}{2}mathbf{1}_{xi<0}+frac{5}{3}mathbf{1}_{0leqxileq2}+9mathbf{1}_{xi>2}$$

edit:

Observe that $mathbb P(xi<1midmathcal D)=mathbb E(mathbf1_{xi<1}midmathcal D)$.

We can find it exactly as above if we replace $xi^2$ there by $mathbf1_{xi<1}$:

• $mathbb{E}left[mathbf1_{xi<1}mathbf{1}_{xi<0}right]=aPleft(xi<0right)$




• $mathbb{E}left[mathbf1_{xi<1}mathbf{1}_{0leqxileq2}right]=bPleft(0leqxileq2right)$




• $mathbb{E}left[mathbf1_{xi<1}mathbf{1}_{xi>2}right]=cPleft(xi>2right)$

Or equivalently:

• $a=mathbb P(xi<1midxi<0)=1$


• $b=mathbb P(xi<1mid0leqxileq2)=frac13$


• $c=mathbb P(xi<1mid xi>2)=0$

So we end up with: $$mathbb{P}left[xi<1midmathcal{D}right]=mathbf{1}_{xi<0}+frac{1}{3}mathbf{1}_{0leqxileq2}$$

First, we define the underlying probability space $(Omega, mathcal{A}, mathbb{P}) := ({-2,...,3}, 2^{{-2,...,3}}, mathrm{Unif}{-2,...,3})$. $xi$ is a random variable from $Omega$ into itself, say $xi equiv mathrm{id}_Omega$. Hence, $xi(omega) = omega$.

Note that $mathbb{E}[xi^2|mathcal{D}]$ is also a random variable, from $Omega$ into itself. Hence, we need to compute its value (in principal) for every $omega in Omega$. Also note that the outcome really depends on the definition of $xi$ above. Only the distribution of $mathbb{E}[xi^2|mathcal{D}]$ is unique.
We can now just apply the formula:
begin{align}
mathbb{E}[xi^2|mathcal{D}](-2) &= frac{(-2)^2 cdot 1/6 + (-1)^2 cdot 1/6}{1/3} = 2.5\
mathbb{E}[xi^2|mathcal{D}](-1) &= frac{(-2)^2 cdot 1/6 + (-1)^2 cdot 1/6}{1/3} = 2.5\
mathbb{E}[xi^2|mathcal{D}](0) &= frac{0^2 cdot 1/6 + 1^2 cdot 1/6 + 2^2 cdot 1/6}{1/2} = 5/3\
mathbb{E}[xi^2|mathcal{D}](1) &= frac{0^2 cdot 1/6 + 1^2 cdot 1/6 + 2^2 cdot 1/6}{1/2} = 5/3\
mathbb{E}[xi^2|mathcal{D}](2) &= frac{0^2 cdot 1/6 + 1^2 cdot 1/6 + 2^2 cdot 1/6}{1/2} = 5/3\
mathbb{E}[xi^2|mathcal{D}](3) &= frac{3^2 cdot 1/6}{1/6} = 9.
end{align}
If I didn't do any mistakes, this should be the correct conditional expectation, given the modelling of $xi$.
The second example you can probably solve yourself, remember that $mathbb{P}(xi < 1|mathcal{D} ) = mathbb{E}[mathbf{1}_{xi < 1}|mathcal{D}]$.