New answers tagged conditional-expectation
2
votes
Joint density function: $f_{X,Y}(x,y) = \frac{x^3}{2} e^{-x(y+1)}$. Find $\mathbb{E} (Y|X)$, $\mathbb{E} (Y^2|X^2)$ and $\mathbb{P} (Y>1|X^3 + 1)$.
Generate $X\sim\text{Gamma}(k=3, \theta=1)$, and $Y\vert X=x\sim \text{Exponential}(\lambda=x).$
You can check that this gives the right joint density:
$$f(x,y)=f_x(x)f_{y|x}(y\vert x) \propto (x^{3-1}...
- 1,575
1
vote
Accepted
Notation of Conditional Expectation
I think you mean that $\mathbb{E}\left(X \, \middle|\, Y,Z\right)$ is $\sigma(Y,Z)$ - measurable, not $\sigma(X,Y)$ - measurable.
The notation $\mathbb{E}\left(X \, \middle|\, Y,Z=z\right)$ or $f(Y,z)$...
- 583
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How to prove that $\forall a\in\mathbb R^+: \frac{\int_a^\infty x\cdot e^{-x^2/2}dx}{\int_a^\infty e^{-x^2/2}dx}\le a + \sqrt{2/\pi}$?
We have
$$\int_a^\infty x\mathrm{e}^{-x^2/2}\,\mathrm{d} x = \mathrm{e}^{-a^2/2}.$$
It suffices to prove that
$$f(a) := \int_a^\infty \mathrm{e}^{-x^2/2}\,\mathrm{d}x
- \frac{\mathrm{e}^{-a^2/2}}{a + \...
- 30.2k
1
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Why is this L2 a (closed) subspace of another L2?
The author means that $ X = L^2(\Omega, \mathcal{F}, P) $ is a subspace of $ Y = L^2(\Omega, \mathcal{G}, P) $ in the sense that there is a natural inclusion $ \iota : X \hookrightarrow Y $. Let $ \...
- 3,042
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Bound on expectation of minimum of two stopping times (proving integrability)
I will give a sketch for people who stumble on this question. (since this is a past paper question for Part III Advanced Probability, there will probably be plenty more people landing here in the ...
- 1,439
1
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Equality in distribution of a ratio of independent RVs
I'm not sure how usual it is to leave a hint on a three year old question, but I do this for the benefit of people who might stumble upon this question.
Let $U$ be a uniformly distributed random ...
- 1,439
0
votes
Accepted
Equilibrium values in test-taking game
I think the issue might be in $P(Y = 1|X=1) = P(Y = 1| \theta,X=1) = \theta$.
Lets assume $P(Y = 1| \theta,X=1) = \theta$ and proceed.
By marginalization we have, $$P(Y = 1 | X = 1) = \int_0^1 P(Y = 1,...
- 2,920
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Is there a quick trick to see $\mathbb{E}[\theta|L=l] = \mathbb{E}_{\Lambda}\left[ \mathbb{E}[\theta|\Lambda, L=l] | L=l \right]$ without expanding?
By the tower rule:
$$\mathbb{E}[\theta\mid L] =\mathbb{E}\Big[\mathbb{E}[\theta\mid\Lambda, L]\mid L\Big]$$
From where:
$$\mathbb{E}[\theta\mid L=l] = \mathbb{E}\Big[ \mathbb{E}[\theta\mid\Lambda, L]\...
- 153
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In linear regression, is there a formal argument that fixed design is "equivalent" to regressors being independent of random errors?
These approaches are computationally "equal" if all the calculations (derivations) for the second approach, i.e., $\{(Y_i, X_i)\}_{i=1}^n$, are done by conditioning on $X=x$. Otherwise, they ...
- 16.1k
2
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Accepted
If $E|Y|\lt \infty$, $E[Y|X] = m(X)$ and $(X_1,Y_1)$ come from same distribution as $(X,Y)$. Is it true that $E[Y_1|X_1] = m(X_1)$?
It's true. Let $B$ be a Borel subset of the real line. Then $$\int Y \mathbb 1(X\in B)\,dP = \int y \mathbb 1(x \in B)\,\mu(dx,dy) = \int m(x)\mathbb 1(x\in B)\,\mu(dx,dy)$$
, where $\mu$ is the law ...
- 857
1
vote
Accepted
Show that $Var(X| \mathcal{G}) \leq Var(X)$
This is not true.
For a counterexample, suppose $X$ takes values $-1, 0, 1$ each with probability $1/3$ and let $\mathcal{G} = \sigma(\{X=0\})$.
Then on $\{X\neq 0\}$, $$\operatorname{Var}(X\mid \...
- 17.5k
1
vote
Accepted
Conditional Expectation of product, where one conditional expectation is already known.
Short answer No. Take $Z = X$ and non-$\mathcal G$ measurable then,
$$\mathbb E\left[XZ\Big | \mathcal G\right] = \mathbb E \left[X^2 \Big |\mathcal G\right] \neq \mathbb E \left[X \Big |\mathcal G\...
- 7,696
2
votes
Accepted
Expectation Conditioned on $\sigma$-subalgebra
Yes, it is certainly fine to have another variable to index the union in your expression for $\sigma(X)$. The tricky part about trying for such an explicit expression for $\sigma(X)$ is that it's ...
- 11.5k
2
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Two random variables i.i.d and same distribution function, compute $P(X\leq Y|Y)$
$\mathbb P\left[X \le Y |Y\right]$ is the (random) probability of $X$ is smaller than $Y$ knowing the (random) value of $Y$. For instance if $X$ and $Y$ are independant, $$\mathbb P\left[X \le Y |Y\...
- 7,696
1
vote
Accepted
Is the conditional expectation function of a convex function convex?
You are right, take $g(x,u)=x^2-2u^2$ and suppose that $X=U$. Then $g(\,\cdot\,, u)$ is convex for all $u$ and $m(x)=\mathbb E[g(X,U)|X=x]=g(x,x)=-x^2$ which is not convex.
- 5,043
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Conditional expectation given event from first principles
I actually ended up figuring it out for myself. I write the answer as it may be of use to others. The more elementary approach is to exploit Radon-Nikodym theorem, due to the absolute continuity of $\...
- 768
1
vote
Accepted
Understanding the dependence of conditional probability on filtrations
This is generally true if $H$ is an atom of $\mathcal F_n$, i.e. if $G \in \mathcal F_n$ and $G$ is a strict subset of $H$ implies $\mathbb{P}(G)=0$. Otherwise, if $H = H_1 \cup H_2$, it would imply $...
- 11.5k
2
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Issue with Application of Law of Total Probability to Expectation of a Geometric Random Variable
I have managed to find the answer myself. My assumption that
$$
Z|F \sim Geom\left(\frac{p_A}{1 - p_B}\right) \\
Z|F^c \sim Geom\left(\frac{p_B}{1 - p_A}\right)
$$
was wrong. In fact, it can be shown ...
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