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9

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


8

Consider two normal distributions with the same variance and different means.


8

Use the strong law of large numbers. Choose $B\in M$ so that $\lambda(B)\neq \mu(B)$, and consider the disjoint sets $$\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \lambda(B)\right\}\mbox{ and }\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \mu(B)\right\}.$$


7

Let $X=Z$, and $Y$ such that $P(X \le Y) = \frac{1}{2}$ and $P(X=Y)=0$, then in order for the equality to hold we must have \begin{eqnarray} 0 &=& P(X=Y) \\ &=&P(X \le Y \le Z) \\ &=& P(X \le Y) P(Y \le X)\\ &=& P(X \le Y) \left( 1- P(X<Y) \right) \\ &=& \frac{1}{4} \end{eqnarray} So the equality does not hold in ...


5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ...


5

Assuming at least that $\Bbb E|X_j|<\infty$: Clearly yes if $\sum|a_j|<\infty$; then Chebyschev says that $\sum P(|a_jX_j|>\epsilon)<\infty$ for every $\epsilon>0$. No in general. Assuming $X_j$ is not essentially bounded there exist $a_j\to0$ such that $\sum P(|a_jX_j|>1)=\infty$, so the less trivial half of Borel-Cantelli says that ...


5

I'll assume real scalars for convenience. If $y = \int_D f\; dP \notin C$, then by the Separation Theorem there is $r \in \mathbb R$ and a continuous linear functional $\phi$ such that $\phi(y) > r$ while $\phi \le r$ on $C$. But $\phi(y) = \int_D \phi(f)\; dP \le \int_D r \; dP = r$, contradiction.


4

Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$. This sort of thing is usually how filtrations are created.


4

By the definition of the conditional expectation, $$\mathbb{P}(X_{n+m} = i \mid X_n = j, X_0 = j_n) = \frac{\mathbb{P}(X_{n+m}=i, X_n = j, X_0 = j_n)}{\mathbb{P}(X_n = j, X_0 = j_n)}. \tag{1} $$ Now, by the Markov property (MP) $$\begin{align*}& \mathbb{P}(X_{n+m}=i, X_n = j, X_0 = j_n) \\ &= \sum \mathbb{P}(X_{n+m} = i, X_{n+m-1} = ...


4

$$F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(X\leq x)$$ $$F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(Y\leq y)$$ $$F_{X,Y}(x,y)^2\leq P(X\leq x) P(Y\leq y)$$


4

Probability is not just defined on a set. It's also defined on a $\sigma$-algebra of subsets. Namely, a probability structure is made of three parts: The underlying set, say $X$. The sets to which you can assign probability. The function which assigns probability. There are several axioms which we require of these structures, such as the sets to which ...


4

If you wonder why so far only the "trivial" example of a Dirac measure has been mentioned, there is actually a good reason for this. The young Ulam wondered whether there is an uncountable set $X$ for which there is a probability measure $$ \mu \colon \mathcal P(X) \rightarrow [0,1] $$ that vanishes on singletons (i.e. $\mu(\{x\})= 0$ for all $x \in X$). He ...


4

Let $(X_j)_{j \in \mathbb{N}}$ be a sequence of independent random variables such that $\mathbb{P}(X_j = 1) = \mathbb{P}(X_j =-1) = \frac{1}{2}$. The event $\{X_j = 1\}$ models that the $j$-th child is a boy, $\{X_j = -1\}$ models the event that it is a girl. Then $$S_n := \sum_{j=1}^n X_j$$ describes how much more boys are born than girls until the $n$-th ...


4

Actually, something much stronger is true: If $P$ is a measure on a $\sigma$-algebra $\mathcal{F}$ and $\mathcal{G}\subseteq\mathcal{F}$ is a $\sigma$-subalgebra on which $P$ is atomless and $\sigma$-finite, then $P$ is atomless on all of $\mathcal{F}$. To show this, suppose $A\in \mathcal{F}$ is an atom. Then for any $B\in \mathcal{G}$, either $P(A\cap ...


4

Hint: Consider $(X,Y) := (X,X)$ for a continuous random variable $X$. Show that $$D := \{(x,y) \in \mathbb{R}^2; x=y\}$$ has Lebesgue measure zero, but $\mathbb{P}((X,Y) \in D)=1$. Conclude that $(X,Y)$ is not absolutely continuous (with respect to the Lebesgue measure).


3

Your professor is right. The point is that the statement: "a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ and $P(X_n=0)=1-\dfrac{1}{n}$" is not enough to define a sequence of random variables. You need to PROVE that there is "a sequence $\{X_n\}$ of independent random variables defined by: $P(X_n=n)=\dfrac{1}{n}$ ...


3

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ...


3

Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation. Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on $(M, ...


3

The idea is to construct a measure $\mu$on an product space $\Bbb{R}\times \Bbb{R}$ such that $\mu((-\infty,a),\Bbb{R}) = F(a)$ and $\mu(\Bbb{R},(-\infty,b)) = G(b)$ such that $\mu((x,y): x>y ) = 0$. Once you have this, then $Y = X + Z$ with $Z\geq 0$ and noting that $\phi$ is increasing $$\Bbb{E}[\phi(Y)] = \Bbb{E}[\phi(X+Z)] \geq \Bbb{E}[\phi(X)] ...


3

The answer given by @d.k.o. is almost correct: Let us assume that $\int_{A_n} X \, dP \not \to 0$. Then there is some $\epsilon > 0$ and a subsequence $A_{n_k}$ with $|\int_{A_{n_k}} X \, dP | \geq \epsilon$ for all $k$. Now, since $P(A_n) \to 0$, we can choose a further subsequence $A_{n_{k_\ell}}$ (which we call $B_\ell$ for brevity) with $P(B_\ell) ...


3

To flesh out Did's comment: let $P(X=1)=P(X=-1)=1/2$. Let $Y=-X$ and note that $X,Y$ are not independent. Then define $f(x,y)=x+y$. $0=E[X+Y|Y=1]\neq E[X+1]=1.$


3

$P(X\leq Y)=\int_{y}P(X\leq y)dF_Y(y)\geq \int_yP(Y\leq y)dF_Y(y)=\int_{-\infty}^\infty F_Y(y)dF(y)=\dfrac{1}{2}$. So the answer is NO, in general. You can only guarantee that $P(X\leq Y)\geq\dfrac{1}{2}$. For example, consider $X\in N(0,1)$ and $Y\in N(1,1)$ then check that $P(X\leq x)\geq P(Y\leq x)$ for all $x\in\mathbb R$. However, also check that ...


3

The means add regardless of any other assumptions. Assuming (as usual in random walks) that the increments are independent, the variances also add. Hence the variance at time $t$ is $\sigma^2t$, and the mean is $\mu t$. Now $$\text{Var}(x_t)=E[x_t^2]-(E[x_t])^2=E[x_t^2]-\mu^2t^2=\sigma^2t$$ hence $$E[x_t^2]=\sigma^2t + \mu^2t^2.$$


3

We have $X,Y$ real valued with $Y(\omega)\leq X(\omega)\Leftrightarrow0\leq X-Y$ for all $\omega\in \Omega$ and $\mathbf{E}(X),\mathbf{E}(Y)$ exists, we set as mentioned in the comments $$ Z=X-Y\geq0 $$ Now we just plug in and get $$ ...


3

The sum of independent normal random variables is normal. That assumption of independence is very important: don't leave it out! The mean of the sum is the sum of the means (expected value is always additive), the variance is the sum of the variances (variance is additive for independent random variables).


3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...


3

Let $Z=X+Y$. We know $0\leq Z\leq 3$ but for the density of $Z$ we have three cases to consider due to the ranges of $X$ and $Y$. If $0\leq z\leq 1$: \begin{eqnarray*} f_Z(z) &=& \int_{x=0}^{z} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=0}^{z} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=0}^{z} \\ &=& 2z - ...


3

$$E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right] = E\left[E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \right] \right] = ...


3

As I mentioned in my comment, there is an analogue of this for $\sigma$-fields but it is a lot more complicated. To slightly simplify the discussion, I'm going to pretend the only operations we care about are union and intersection, and ignore complements. In fact, because of the identities $(\bigcap A_i)^c=\bigcup A_i^c$ and $(\bigcup A_i)^c=\bigcap ...


3

The CDF of $Y_n=\max_{1\le i\le n}\{v_i\}$ is $$F_{Y_n}(t)=[F_v(t)]^n\rightarrow 1\{t=\bar v\}\text{ as }n\rightarrow \infty$$ which means that $Y_n\xrightarrow{d} \bar v$.Using the a.s. representation there is a sequence $\{Y_n'\}$ where each $Y_n'$ has the same distribution as $Y_n$ and $Y_n'\xrightarrow{a.s}\bar v$. Since $|Y_n'|\le |\underline ...



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