Finding conditional expectation and conditional probability

Question

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?

Answer 1

I assume that $\mathbb E[\xi^2\mid\mathcal D]$ denotes the same as $\mathbb E[\xi^2\mid\mathcal\sigma(\mathcal D)]$

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$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]$ is measurable wrt $\sigma\left(\mathcal{D}\right)$ implying here the existence of coefficients $a,b,c$ with $$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]=a\mathbf{1}_{\xi<0}+b\mathbf{1}_{0\leq\xi\leq2}+c\mathbf{1}_{\xi>2}\tag1$$

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

Substituting $\left(1\right)$ in $(2)$ we find for $D=\{\xi<0\}$, $\{0\leq\xi\leq2\}$ and $\{\xi>2\}$:

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

• $\mathbb{E}\left[\xi^{2}\mathbf{1}_{0\leq\xi\leq2}\right]=bP\left(0\leq\xi\leq2\right)$

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

Or equivalently

• $a=\mathbb{E}\left[\xi^{2}\mid\xi<0\right]=\frac{1}{2}\left(-2\right)^{2}+\frac{1}{2}\left(-1\right)^{2}=\frac{5}{2}$
• $b=\mathbb{E}\left[\xi^{2}\mid0\leq\xi\leq2\right]=\frac{1}{3}0^{2}+\frac{1}{3}1^{2}+\frac{1}{3}2^{2}=\frac{5}{3}$
• $c=\mathbb{E}\left[\xi^{2}\mid\xi>2\right]=3^{2}=9$.

So we end up with: $$\mathbb{E}\left[\xi^{2}\mid\mathcal{D}\right]=\frac{5}{2}\mathbf{1}_{\xi<0}+\frac{5}{3}\mathbf{1}_{0\leq\xi\leq2}+9\mathbf{1}_{\xi>2}$$

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edit:

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

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

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

• $\mathbb{E}\left[\mathbf1_{\xi<1}\mathbf{1}_{0\leq\xi\leq2}\right]=bP\left(0\leq\xi\leq2\right)$

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

Or equivalently:

• $a=\mathbb P(\xi<1\mid\xi<0)=1$
• $b=\mathbb P(\xi<1\mid0\leq\xi\leq2)=\frac13$
• $c=\mathbb P(\xi<1\mid \xi>2)=0$

So we end up with: $$\mathbb{P}\left[\xi<1\mid\mathcal{D}\right]=\mathbf{1}_{\xi<0}+\frac{1}{3}\mathbf{1}_{0\leq\xi\leq2}$$

Answer 2

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 $\mathbb{P}(\xi < 1|\mathcal{D} ) = \mathbb{E}[\mathbf{1}_{\xi < 1}|\mathcal{D}]$.