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3

Hints: Since $(B_t)_{t \geq 0}$ is a martingale it follows from the optional stopping theorem that $$\mathbb{E}(B_{T_R \wedge T_0}) = x.$$ Deduce from $B_{T_R \wedge T_0} \in \{0,R\}$ that $$\mathbb{P}(B_{T_R \wedge T_0}=R) = \frac{x}{R} \qquad \mathbb{P}(B_{T_0 \wedge T_R} = 0) = \frac{R-x}{R}. \tag{1}$$ Conclude that \mathbb{P}(T_R<T_0) = \frac{x}{R}.... 2 You can directly compute the expectation of Z as follows: \begin{align*} \mathbb E[\min(X,Y)]&=\int_{400}^{800}\int_{500}^{600}\min(x,y)\dfrac{\mathrm{d} x}{100}\dfrac{\mathrm{d} y}{400}\\ &=\dfrac{1}{40000}\left(\int_{400}^{500}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y + \int_{500}^{600}\int_{500}^{600}\min(x,y) \ \mathrm{d}x\mathrm{d}y +... 2 Since stopping times typically depend on a path of a process (and not only the value at some particular time), your formula for the conditional expectation is not too useful. However, there is a " functional" version of this formula which comes in handy; you can find it for instance in the book Brownian motion - an introduction to stochastic processes by ... 2 Hint: When faced with a random variable X and an event A, the following may help:E[X|A] = \frac{E[X1_{A}]}{P(A)}$$where 1_{()} is the indicator function. In your case replace X with f(X,Y) and A with \{y \in \mathcal{Y}\}. Let me know if you need more help. 2 You have  \mathbb V (X- \mathbb E[X|Y]) = \mathbb E[ (X-\mathbb E[X|Y])^2]. The conditional variance is  \mathbb V( A|B) := \mathbb E ( (A - \mathbb E [A|B])^2 | B ) . what you need is the "tower rule/law of total expectation"[wiki]$$ \mathbb E\big[\mathbb E[A|B]\big] = \mathbb E [A]$$which means that$$ \mathbb E[ (X-\mathbb E[X|Y])^2] = \mathbb E\...

2

$$\mathbb E\left[X_t\int_s^t Y_r\,\mathrm d r\mid \mathcal F_s\right]-\mathbb E\left[\int_s^t X_rY_r\,\mathrm d r\mid \mathcal F_s\right]=\mathbb E\left[\int_s^t Y_r(X_t-X_r)\,\mathrm d r\mid \mathcal F_s\right]$$$$\underset{(*)}{=}\int_s^t\mathbb E\big[Y_r(X_t-X_r)\mid \mathcal F_s\big]\,\mathrm d r.$$ A reference for $(*)$ is given here. Using the fact ...

2

It seems to be true. Here are some steps towards a proof. We first treat the case where $F_{13}\in\mathbb F_1$. Define for a fixed $G_1\in\mathbb F_1$ the collection of sets $$\mathcal A:= \{A\mid \Pr\left(G_1\cap A\mid\mathbb F_3 \right)=\Pr\left(G_1 \mid\mathbb F_3 \right)\Pr\left( A\mid\mathbb F_3 \right) \mbox{ a.s.}\}.$$ We can check that $\mathcal A$...

2

If $f_n \to f$ almost everywhere define $g(x)=\lim f_n(x)$ if the limit exist and $0$ otherwise. Then we can show that $g$ is measurable and $f=g$ a.e.. Hence $f$ and $g$ are equal as elements of $L^{2}$. Completeness of the measure space is not required for this argument.

1

First of all, we notice that \begin{align} E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] &= E[X_tY_t - X_tY_s -X_sY_t +\underbrace{X_sY_s}_{\mathcal{F}_s-\text{mesurable}}|\mathcal F_s]\\ &= E[X_tY_t|\mathcal F_s] - Y_sE[X_t|\mathcal F_s] - X_sE[Y_t|\mathcal F_s] + X_sY_s \\ &= E[X_tY_t|\mathcal F_s] - Y_sX_s - X_sY_s + X_sY_s \\ &=E[X_tY_t-...

1

A good thing about (zero-mean) multivariate normal distribution is that its structure is compatible with the inner-product structure arising from its covariance structure: \begin{array}{ccc} \text{zero-mean gaussian distribution} & & \text{inner product space} \\ \hline \text{covaraince} & \leftrightarrow & \text{inner ...

1

Let's be more careful here about the space of functions ($\mathscr{L}^2$) and the space of equivalence classes of functions ($L^2$). Fix a sequence $X_n\in\mathscr{L}^2(\Omega,\mathcal{B},\mathbb{P})$ of random variables that converges in the seminormed space $\mathscr{L}^2(\Omega,\mathcal{A},\mathbb{P})$ to some $X$. Then by extracting a subsequence, we ...

1

Hints: it can be shown that the Borel sigma algebra on $C_0$ is the smallest sigma algebra which makes the evaluation maps $f \to f(t)$ ($t \in \mathbb R$) measurable. From this it follows immediately that $\omega \to (W_t(\omega))_t\geq 0$ is a measurable function from $\Omega$ with the sigma algebra $\sigma((W_t): \geq 0)$ into $C_0$. Hence $F(W_t(\omega))... 1 Observe that $$E[E^{(i)}[ZT]]=E[ZT]$$ hence it suffices to check that $$E\left[ E^{(i)}[Z]E^{(i)}[T]\right]= E\left[ ZE^{(i)}[T]\right].$$ Denoting$\mathcal F$as the$\sigma$-algebra generated by$X_j$,$j\in\{1,\dots,n\}\setminus \{i\}$, we get that $$E\left[ E^{(i)}[Z]E^{(i)}[T]\right]=E[E[Z\mid \mathcal F]E[T\mid \mathcal F]]$$ and since$E[...

1

For suitable $g:\Omega_Y\to\mathbb R$ we find: \begin{aligned}\mathbb{E}\left[g\left(Y\right)\int f\left(x,Y\right)P^{X}\left(dx\right)\right] & =\int g\left(y\right)\int f\left(x,y\right)P^{X}\left(dx\right)P^{Y}\left(dy\right)\\ & =\int\int g\left(y\right)f\left(x,y\right)P^{X}\left(dx\right)P^{Y}\left(dy\right)\\ & =\int\int g\left(y\... 1 So, you want to show that\mathbb{E}_\mathbb{Q} \left|{1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}\left[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G\right]\right| < \infty.$$By the conditional triangle inequality, it is enough to show$$\mathbb{E}_\mathbb{Q} \left[{1 \over \mathbb E_{\mathbb P}[{{...

1

No you can't. Let $Y,Z$ be independent copies of the two state symmetric Markov chain starting with the invariant distribution. Let $X=1_{Y=Z}-1_{Y\neq Z}$. It is easy to check $$\mathbb{E}[X\mid Y]=\mathbb{E}[X\mid Z]=0,$$ but of course $X\neq 0$.

1

Let $(a,b)$ be a pair of cards in your deck. The probability that $(a,b)$ is a black-black pair is $25/102$: as @lulu pointed it out $$\mathbb P((a,b) \ \textrm{black})=\mathbb P(a \ \textrm{black}) \times \mathbb P(b \ \textrm{black} | a \ \textrm{black})=(1/2)\times (25/51)=25/102.$$ Let $X_{(a,b)}$ be the random variable defined as $1$ if $(a,b)$ is a ...

1

Since $H\in \mathcal H\subset \mathcal G$, $$\boldsymbol 1_H\mathbb E[X\mid \mathcal G]=\mathbb E[\boldsymbol 1_HX\mid \mathcal G],\quad \quad \text{and}\quad \boldsymbol 1_H\mathbb E[X\mid \mathcal H]=\mathbb E[\boldsymbol 1_HX\mid \mathcal H].$$ Therefore $$\mathbb E\big[\mathbb E[\boldsymbol 1_HX\mid \mathcal H]\big]=\mathbb E[\boldsymbol 1_HX]=\mathbb E\... 1 By definition of conditional expectation (https://en.wikipedia.org/wiki/Conditional_expectation): \int_{H}\mathbb E[X|\mathcal{H}]dP=\int_{H}\mathbb X dP \int_{H}\mathbb E[X|\mathcal{G}]dP=\int_{H}\mathbb X dP (note that H is both \mathcal{G} and \mathcal{H} measurable) so the two expressions are both equal to the integral of the original function.... 1 Note that E[X|A] is a constant and hence$$ E[E[X|A]1_{A}|B]=E[X|A]\cdot E[1_{A}|B]=E[X|A]P(A|B),  where the first equality follows from the linearity of conditional expectation with respect to an event. You do not need to use independence here.

1

You can view $E\{f(X,Y)|Y=y\}$ as a special case of conditioning on a set. $E\{f(X,Y)|Y=y\}=E\{f(X,Y)|Y\in\{y\}\}$. In general, \begin{align} E\{f(X,Y)|Y\in\mathcal{Y}_1\} &= \sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x, Y=y|Y\in \mathcal{Y}_1)\\ &=\sum_{x\in \mathcal{X}, y\in \mathcal{Y}_1}f(x,y)P(X=x|Y=y)P(Y=y|Y\in \mathcal{Y}_1)\\ &=...

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