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Tim
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Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Update:

Additional Related Question

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Update:

Additional Related Question

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Tim
  • 17
  • 4

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x f_{\mathbb{X}|\mathbb{Y}}(x,y)$$$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x f_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x xf_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

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Tim
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Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conidtionalconditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x f_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conidtional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x f_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

Given a discrete random variable $\mathbb{X}$ on $\Omega = \{1,2,3\}$ with the following pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = \frac{1}{3}$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = \frac{1}{2}$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = \frac{1}{6}$

Find the following value of the conditional expectation: $\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}]$

Conditional Expectation Formula:$$\text{E}[\mathbb{X} | \mathbb{Y} = y] = \sum_x f_{\mathbb{X}|\mathbb{Y}}(x,y)$$

so does:

$$\text{E}[\mathbb{X}\space | \space\mathbb{X} \in \{1,2\}] = 1*\frac{1}{3} + 2*\frac{1}{2} = \frac{4}{3}$$ since $\mathbb{X}$ is conditioned on a subset of itself?

Thanks!

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Tim
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