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2 votes
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Car parking - quant guide

For $1 \le i < j \le 4$, let $(i,j)$ be the situation where the cars are in spaces $i$ and $j$. During the process described in the problem, we will definitely be in the states $(1,2)$, $(1,3)$, $(...
paw88789's user avatar
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1 vote
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Mutual information expansion not justifiable

It's really simple. Think of $(Y,Z)$ as a single multivariate variable $W=(Y,Z)$. Then $$\begin{align}I(X;W) &= E \left[\log \frac{P(X,W)}{P(X)P(W)}\right]\\ &= E \left[\log \frac{P(X|W)}{P(X)}...
leonbloy's user avatar
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1 vote
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Conditional Expectation of dependent normal distribution

We have $X-aZ_1=Y-aZ_2$, so condition both sides with respect to $X$. This gives $$ X -aE(Z_1\mid X) = E(Y\mid X) -a E(Z_2\mid X). $$ By independence we have $E(Z_2\mid X)=E(Z_2)=0$ while $$E(Z_1\mid ...
grand_chat's user avatar
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1 vote

Conditional Expectation of dependent normal distribution

Without loss of generality, assume that $c=0$ and $a=1$. Let us find $K$ such that $$ \operatorname{Cov}\left(Y-KX,X\right)=0. $$ Since $(X,Y)$ is a Gaussian vector, this will imply that $Y-KX$ is ...
Davide Giraudo's user avatar
0 votes

Geometric interpretation of joint normal conditional expectation

Visualize a bivariate normal pdf $f_{X.Y}(x,y)$ as a surface above the $x$-$y$ plane. The volume of the solid trapped between this surface and the $x$-$y$ plane is $$\int_{-\infty}^\infty\int_{-\infty}...
Dilip Sarwate's user avatar
2 votes

Expected number of bonus wins for a specific slot machine game

I'm getting the answer $$ \mathbb{E}[B] = \frac{643865593}{215233605} = 2.991473348225524 $$ My idea is to use a Markov chain with states $(a,b)$ for $0\leq a,b \leq 8$ where the states where $8$ ...
ploosu2's user avatar
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2 votes
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Expected number of original items after some replacements

The first item will continue to be the original after $n$ weeks iff in all weeks only items different than the first are selected. What is the chance it wasn't selected in week $1$? In week $2$? In ...
JMoravitz's user avatar
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1 vote
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Derivation of a conditional expectation

Your formula only holds for a normally distributed $\varepsilon$, not for an arbitrary distribution with finite variance, so there won't be a proof independent of distribution. Suppose $\varepsilon$ ...
Zoe Allen's user avatar
  • 4,798
2 votes

Time-dependent transition probabilities

Here is an alternate proof that doesn't use the hint. Label the cigars in order that they get selected for the first time. Let $E_i$ be the event where the $i$th cigar gets selected a second time ...
mercio's user avatar
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2 votes
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Time-dependent transition probabilities

Your recurrence relation is correct (apart from a small misprint) and being supplemented with the boundary condition $v_n(0)=n$ is in fact solved by the expression given as a hint. Your expression: $$ ...
user's user avatar
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1 vote
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Expectation of the process adapted to the filtration of the Wiener process

The last line in (3) is incorrect. $E[\sigma_t|W_t]$ is a random variable itself depending on $W_t$, you cannot take it out of the outer expectation and you cannot split the expectation linearly ...
gabalz's user avatar
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2 votes
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Let $(x_t: t\geq 0) $ be the Brownian motion. $E(f| x_t)=0$ implies $f=0$?

Corrected answer: Write $x=(x_t: t\geq 0)$ and define $f(x):=(2x_1-x_2)(x_3-x_2)$. Consider the two sigma-fields $G=\sigma(x_t: t\le 2)$ and $H=\sigma(x_t: t \ge 2)$. Clearly $E(f|G)=(2x_1-x_2)E(x_3-...
Yuval Peres's user avatar
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0 votes

Representing a conditional expectation

I will expand on Sangchul's comment that the solution depends on how we define $\mathbf{E}\left( X \mid X \in B \right)$, where $B$ is any Borel set. Suppose we work in a probability space $\left( \...
msantama's user avatar
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3 votes
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Representing a conditional expectation

The solution hinges on how $\mathbf{E}[\cdot \mid X > c]$ is defined. If it is defined as the expectation with respect to the conditional probability measure $\mathbf{P}(\cdot \mid X > c)$, then ...
Sangchul Lee's user avatar
2 votes

Representing a conditional expectation

By definition of $E[X\mid X>c]$: $$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$ And by definition of conditional expectation: $$E[X1_{X>c}]=E\big[E[X\mid\sigma(...
Speltzu's user avatar
  • 583
2 votes

Decomposing a general stopping time into stopping components

(The $y$ in Frid's notation is misleading.) Instead, for a stopping time $\tau\ge 1$ and a state $x$, define $T_x(\omega):=\tau(x\omega)-1$. (Here, for a path $\omega=a_0a_1a_2\cdots$, the ...
John Dawkins's user avatar
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1 vote
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Calculating a Conditional expectation

But this only left me with $$E[X_i|X_{\max}]=X_{\max}\mathbb{P}(X_i=X_{\max}|X_{\max})+E[X_i\mathbf{1}_{\{X_i<X_{\max}\}}|X_{\max}],$$ which doesn't really gelp me I think. Mixing the notation ...
Graham Kemp's user avatar
2 votes

Calculating a Conditional expectation

Let $V=\max\{X_2,\ldots,X_n\}$. For $0\leq v\leq \lambda$ we have $$ \begin{eqnarray} F_V(v) &=& \Pr(V \leqslant v) = \Pr(\max\{X_2,\ldots,X_n\} \leqslant v) \\ &=& \Pr(X_2 \leqslant ...
Julio Puerta's user avatar
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0 votes

Conditional expectation of symmetric statistic has the same distribution when conditioning on iid random variables

In general, if $(X,Y)$ and $(X',Y')$ are identically distributed so are $E[X|Y]$ and $E(X'|Y']$. In this case, by the symmetry of $S$, $(S(X_1,...,X_n),X_i)$ and $(S(X_1,...,X_n),X_j)$ are.
Speltzu's user avatar
  • 583

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