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4

Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$ Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}. $$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$. So the answer is "no" to that main question. It ...


3

I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of ...


3

An alternative solution $$ \begin{align} E(X^r) & = \int_0^{\infty} r x^{r-1} P(X > x) \:dx \\ &= \int_0^{\infty} r x^{r-1} \left[ \int_{y=x}^{\infty} f(y) \:dy \right] \:dx \\ &= \int_{y=0}^{\infty} f(y) \left[ \int_{x=0}^{y}r x^{r-1} \,dx \right]\:dy \\ &= \int_0^{\infty} y^r f(y)\,dy \end{align} $$


3

The total number of ways to arrange $14$ people is $14!$. The number of ways to arrange $14$ people with a specific person last is $13!$. So in a single experiment, the probability of that person being chosen last is $\frac{13!}{14!}=\frac{1}{14}$. Hence in $3$ experiments, the probability of that person being chosen last every time is ...


3

Informally, if we know that $X_1$ is small in absolute value, then $X_2$ cannot be small in absolute value. More formally, let $A$ be the event $|X_1|\le \frac{1}{2}$. A look at the sine curve shows that $\Pr(A)=\frac{1}{3}$. Similarly, let $B$ be the event $|X_2|\le \frac{1}{2}$. We have $\Pr(B)=\frac{1}{3}$. But since $X_1^2+X_2^2=1$, the event $A\cap ...


2

Here is a guideline. Some details are left as smaller exercises, but I really encourage you to fill them up. I'll first recall how to prove that a sequence of random variables converges almost surely, before adapting the argument to the limsups. Convergence of sequences of random variables Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and ...


2

For (i): Suppose for a contradiction that $P(A\cap T^{-n}(A))=0$ for all $n\geq 1$, and consider the sets $A_n=T^{-n}(A)$. Since $P(A)>0$ and $T$ is a measure-preserving transformation, $P(A_n)=P(A)>0$. On the other hand, note that for $m<n$, \begin{align*} P(A_m\cap A_n)&=P(T^{-m}(A)\cap T^{-n}(A))=P(T^{-m}(A\cap T^{-(n-m)}(A)))\\ ...


2

Note that $$ \left( \frac{N_t}{t}-\lambda \right)^2= \frac{1}{t^2} (N_t-t\lambda)^2 \leq \frac{1}{\sigma^2} (N_t-t\lambda)^2$$ for all $t \in [\sigma,\tau]$. Hence, $$\mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} \left( \frac{N_t}{t}-\lambda \right)^2 \right] \leq \frac{1}{\sigma^2} \mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} (N_t-t\lambda)^2 \right].$$ ...


2

The answer depends on your choice of parametrization. The way you have characterized the survival function $$S_X(x) = \Pr[X > x] = \exp(-\alpha x^{\beta - 1}), \quad x \ge 0,$$ is a little different than other parametrizations. Without seeing the source document to determine how the authors themselves chose to parametrize the distribution, it is ...


2

A nonnegative random variable $X$ is exponentially distributed with parameter $\lambda$ if $$P[X\geq t]=e^{-\lambda t}\qquad(t\geq0)\ .$$ It has a probability density $$f_X(t):=\lim_{h\to0+}{P[t\leq X\leq t+h]\over h}=\lim_{h\to0+}{e^{-\lambda t}-e^{-\lambda(t+h)}\over h}=\lambda\>e^{-\lambda t}\ .$$ Your last displayed formula then should read ...


2

Two events $A$ and $B$ are independent if and only if their joint probability equals the product of their probabilities: $$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$$ Here we have that $$P(E_1 \cap E_2) = \frac{1}{36}$$ $$P(E_1) \cdot P(E_2) =\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$ So they are independent. On the other hand $$P(E_1) = ...


2

There are two schemes to define the the integral $\int_{\Omega}X(\omega)dP(\omega)$, for $X$ a random variable taking values in a Banach space --- Bochner integral and Pettis integral --- depending on the strong or weak measurability of $X$ respectively. Click here for Bochner integral, and here for Pettis integral, both from Wikipedia.


2

In the first integral of the third line, you should write $(Y-X)^+$, since you only integrate when $Y\geq X$. Similarly, then the integrand in the second integral becomes $(X-Y)^+=(Y-X)^-$, and then you have $E(Y-X)^+-E(Y-X)^-=E(Y-X)$.


2

When discussing samples, there is a vector-valued random variable $Y$ where the real-valued random variables $Y_i$ are independent and have the same distribution as $X$. Then $\{ x_i \}_{i=1}^n$ is $Y(\omega)$ for some particular $\omega$ in the sample space $\Omega$. Sometimes when we refer to a sample we actually mean $Y$ itself rather than a particular ...


2

Let $A_n$ be the natural numbers with the first $n$ natural numbers thrown out. I.e. $A_n=\mathbb N\setminus\{1,2\cdots n \}$. Then each $\mu(A_n)=\infty$ and $\cap A_n=\phi$ so that $\mu(\cap A_n)=0$. Can you adapt this argument to your example?


2

Choose $x_k \in \Omega$ such that each $x_k$ is distinct. Let $A_n = \{x_k\}_{ k \ge n}$.


2

There is a set $A \subset \Omega$ that can be identified as $\mathbb{Z}$ (in the sense that there is a bijection $f: \mathbb{Z} \to A$). Let $A_n = \mathbb{Z} \setminus \left\{ 1 ,\dots, n \right\}$. This sequence satisfies the claim.


2

Try this, which I think is similar to your strategy. Let $C=\{x_1, x_2, \cdots \}$ be a countably infinite set of distinct points. Let $A_n:= \{x_i \in C : i\geq n\}$. So $A_1=C$, $A_2=\{x_2, x_3,\cdots \}$, etc. Then Clearly $$\bigcap_n A_n =\emptyset$$ and $A_n \downarrow$. However, for each $n$, $\mu(A_n)=\infty$, so $\mu(A_n)$ cannot tend to 0.


2

To be true that $A_i \downarrow A \implies \mu(A_i) \downarrow \mu(A)$, you should have that $\mu(A_1) < \infty$. Hint: construct an example where $\mu(A_1) = \infty$.


2

Note that $X$ is a finite set. So given $x\in X$, the set $B_x= \{ S \in \mathcal{F} : x\in S\}$ is finite. Since $\mathcal{F}$ is a $\sigma$-algebra, we have $X\in B_x$ ($B_x$ is not empty) and $$A_x=\bigcap_{S\in B_x} S \in \mathcal{F}$$ Clearly $x\in A_x$, and $A_x$ is the smallest element in $\mathcal{F}$ containing $\{x\}$. Note that if $y\in X$ and ...


2

This statement is much less scary than it sounds. Recall that a family of probability measures $\mathcal{M} =\{\mu_\alpha \}_\alpha$ on a space $X$ is tight if for every $\epsilon > 0$, we can find a compact set $K_\epsilon \subset X$ so that $$ \mu_\alpha(K_\epsilon) > 1 - \epsilon$$ for every $\alpha$. Let's quickly note that if $X$ is compact, ...


2

Let $C := \bigwedge_{B \in S_1}\bar{B}$ and $D := \bigwedge_{B \in S_2}\bar{B}$. The statement now reads like this: $$\Pr(A\mid C \wedge D) =\frac{\Pr(A \wedge C \mid D)}{\Pr(C \mid D)}.$$ If $\text{Pr}_D(X)$ denotes $\text{Pr}(X \mid D)$, then the statement is equivalent to $$\text{Pr}_D(A\mid C) =\frac{\text{Pr}_D(A \wedge C)}{\text{Pr}_D(C)},$$ which ...


1

Let $D:=\{(t,\omega):f(t,B_t(\omega)\not=0\}$ and $C=\{(t,x):f(x,t)\not=0\}$. Observe that $D=\varphi^{-1}(C)$, where $\varphi(t,\omega) =(t,B_t(\omega))$ (mapping $[0,T]\times\Omega$ into $[0,T]\times\Bbb R$). Consequently, writing $\lambda$ for Lebesgue measure on $[0,T]$, $$ 0=\lambda\otimes\Bbb P(D) =\int\int 1_{C}(t,x){1\over\sqrt{2\pi ...


1

For any measurable set $A \subseteq [0,T] \times \Omega$ it holds that $$\begin{align*} \{(t,\omega); (t,B_t(\omega)) \in A\} &= \int_0^T \! \int 1_A(t,B_t) \, d\mathbb{P} \, dt \\ &= \int_0^T \! \int 1_A(t,x) \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} \, dx \, dt. \end{align*}$$ Since $e^{-x^2/2t}$ is strictly positive, this implies $$(\lambda|_{[0,T]} ...


1

Here's why what you did didn't work. Let $X,Y$ be two independent normally distributed random variables with mean 2 and variance 1 (this is what you sampled from with your R code). This means that $X+Y$ is normally distributed with mean 4 and variance 2, which implies that $$ \frac{X+Y - 4}{\sqrt{2}}$$ is a standard normal random variable, so it follows that ...


1

Hint on i): if $\phi(t)$ is a moment generating function then what can be said about $\phi(0)$? Hints on ii) What are the looks of moment generating function of $cX$ expressed in $\phi_X$ where $c$ is constant? If $X,Y$ are independent then what are the looks of the moment generating function of $X+Y$ expressed in $\phi_X$ and $\phi_Y$?


1

You are correct. You are also correct that the probability of any discrete value from a continuous distribution is zero. In this case you can handle the problem by only looking at the probability of belonging to a small interval around $\bar{z}$, which is nonzero, and can be calculated as follows. $$P(x<X<x+\epsilon) \approx \epsilon f(x)$$ Let ...


1

The first part of your logic is correct.   To find the individual probabilities, you can simply use combinatorics. $$\begin{align}p(9\textsf{ black}) ~=~& \dfrac{\dbinom{50}{9}}{\dbinom{100}{9}}\\[2ex]p(1\textsf{ white}, 8\textsf{ black})~=~&\dfrac{\dbinom{50}{1}\cdot\dbinom{50}{8}}{\dbinom{100}{9}}\end{align}$$ ... and so on. PS: As a ...


1

It depends on the probability model we use, and that is dictated by the physical situation. But if balls are thrown one at a time towards the cells, with all cells equally likely to get the ball, and we have independence, then indistinguishable makes no difference to the probability. So for the second as for the first the probability is ...


1

See Kolmogorov's "Three Series Theorem" if you want the definitive answer on this subject; it gives you results general for all distributions, not just the gamma distribution. https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem



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