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10

The event $\{H,T\}$ is the event that the coin turns up heads or tails. This event always happens and thus has probability $1$. The empty event, $\varnothing$, should not be thought of as the event that the coin lands on its edge. It is assumed that the coin always lands heads or tails. Rather the event $\varnothing$ is the event that there is no outcome, ...


6

Hint: $P( (\cap_i A_i)^C) = P( \cup_i A_i^C) \leq \sum_i P(A_i^C) = \sum_i (1-P(A_i)) = \sum_i (1-1) = 0$.


6

One simple case where your identity is true: Let $Y$ be some nonconstant and positive RV, and let $X:=cY$ for some nonzero $c$. Then $E(X)=cE(Y)=E(X/Y)E(Y)$.


5

First, note that if the Brownian motion is started from $(x_0,y_0,z_0)$, then the probability to hit the cylinder $\{y^2+z^2= \delta^2\}$ before the cylinder $\{y^2+z^2=1\}$ equals$$\frac{\log(y_0^2+z_0^2)}{\log \delta^2}.$$ Let $$ \Theta_{\delta;r}:=\{y^2+z^2\leq \delta^2;x\geq r\}. $$ Stopping the Brownian motion at the first time it hits the sphere of ...


5

Note that $$ \begin{split} F_Z(z) &= \mathbb{P}[Z \le z] \\ &= 1 - \mathbb{P}[Z > z] \\ &= 1 - \mathbb{P}[\min\{X,Y\} > z] \\ &= 1 - \mathbb{P}[X > z, Y > z] \quad \text{now apply independence}\\ &= 1 - (1-F_X(z))(1- F_Y(z)) \\ &= F_X(z) + F_Y(z) - F_X(z)F_Y(z). \end{split} $$ Can you ...


5

Consider $\mathbb{P}(Z > z) = \mathbb{P}(\min\{X, Y\} > z)$. If $\min\{X, Y\} > z$, it follows that $X > z$ and $Y > z$. [This hopefully isn't too difficult to see! If this doesn't make sense to you, grab two numbers. Choose the smallest one. Find a number that this number is greater than (say $k$). Then the other number should be greater ...


5

Hint: Look at $\text{Var}(U)$. It is possible. Ignore the problem at the moment, and consider the function $y = 2$. Does this make sense to you? I hope so, it is a constant, horizontal line at $2$. Similarly, we could have written it as $y = f(x)$. Notice that this means $f(x) =2$. It does not matter that there is no $x$. It still makes sense that it is ...


4

For $k\ge 0$, we calculate the ratio $$\frac{\Pr(X=k+1)}{\Pr(X=k)}.\tag{1}$$ This is $$\frac{e^{-\lambda}\frac{\lambda^{k+1}}{(k+1)!}}{e^{-\lambda}\frac{\lambda^k}{k!}},$$ which simplifies to $$\frac{\lambda}{k+1}.$$ Thus if $\lambda \lt 1$, then the ratio (1) is $\lt 1$ at $k=0$, and even smaller afterwards. Thus $\Pr(X=k)$ is steadily decreasing, and ...


4

The proof below assumes that $X$ and $Y$ belong to the sample space. That is, they map from the sample space to a real number line. Is that also a condition for linearity of expectation? No.   It's the definition of a random variable. Basically a random variable $X$ is a function that maps the sample space to the reals (or a subset there of, called ...


4

Let $X \sim \text{Unif}(\{0,1,2\})$ and $Y=X+1\pmod 3$. Then $X$ and $Y$ are identically distributed, but $(X,Y)$ and $(Y,X)$ are not. For instance, $P((X,Y)=(0,1))=\frac{1}{3}\neq 0 =P((Y,X)=(0,1))$. I'm not sure I understand your second question. What do you mean by "random functions" $\mathbb{R}^2\to \mathbb{R}$, and why should $g((X,Y))=h((Y,X))$ for ...


4

It's a consequence of the axioms of a probability measure: if $A$ is any event, then $A\cup\emptyset=A$ and $A\cap\emptyset=\emptyset$, hence $$ \mathbb{P}(A)=\mathbb{P}(A\cup\emptyset)=\mathbb{P}(A)+\mathbb{P}(\emptyset)$$ Therefore we must have $\mathbb{P}(\emptyset)=0$.


4

Here is a solution that uses SLLN and the additive property of the Poisson distribution. Let $(N_t)_{t \geq 0}$ be a Poisson process of unit rate. Then by the SLLN, together with the inequality $$ \frac{N_{[t]}}{[t]+1} \leq \frac{N_t}{t} \leq \frac{N_{[t]+1}}{[t]}, $$ it is easy to check that $N_t / t \to 1$ as $t \to \infty$ a.s. Now let $T_k = \lambda_1 ...


3

Any subset of the sample space is called an event. If the sample space is finite and has $k$ elements, then there are $2^k$ different events because there are $2^k$ different subsets of a $k$-element set. In your case, $k=2$. Perhaps this might make more sense if we look at the example of rolling a fair die. In this case, the sample space is $S=\{1,2,3,...


3

The right continuity of CDFs is a matter of convention. One could just as well take the default to be the left-continuous function $x\mapsto \Bbb P[X<x]$, as one finds in the Russian literature of yesteryear (and perhaps even today); cf. B.V. Gnedenko's Theory of Probability.


3

When you integrate the conditional density of $X$ given $Y=y$ over all $x$, you should get $1$: $$ \int_{\mathbb{R}}f_{X \mid Y}(x \mid Y=y)dx = 1\tag1 $$ because you've just computed $P(X\in\mathbb{R}\mid Y=y)$. This is true for every value of $y$. So when you attempt to integrate (1) over all values of $y$, you'll be integrating the constant $1$. The ...


3

This is some extension of @jdods's comment. A stochastic process can be understood as a family $X = \{X(t), t\in \mathbb T\}$ of random variables indexed by some parametric set $\mathbb{T}$. Another process $Y = \{X(t), t\in \mathbb T\}$ is a version of $X$ if for all $t\in \mathbb T$ $P(X(t) = Y(t)) = 1$. What is a conditional distribution, or, more ...


3

This question is interesting but to solve it requires to come back at the definitions of conditional distributions, so let us try to be precise. We solve in details a simple case, hoping that this makes apparent the general solution. Assume that $X$ has density $f_X(x)=1$ on $(0,1)$ and that $Y=g(X)$ with $g(x)=6x$ if $x<\frac12$ and $g(x)=4-2x$ if $x&...


3

Suppose, this is not true. Then, we would have $\mu(A\cap\{\lambda_i\ne0\})=0$ for all $i\in\{1,2,\dotsc,l\}$. Hence, $$1_A X_1 = \sum_{i=1}^l \lambda_i1_AY_i$$ implies that $1_AX_1=0$. This, the vector $(1_A,0,\dotsc,0)$ satisfies $$1_AX_1 + 0X_2 + \dotsb + 0X_k = 0$$ which contradicts the linear independence of $(X_1,X_2,\dotsc,X_k)$.


3

In a random sequence of $n$ tosses of a fair coin, the expected total number of runs is $\frac{n+1}{2}$ the expected number of runs of length $L$ or more is $\frac{n+2-L}{2^L}$ when $L \le n$ the expected number of runs of exactly length $L$ is $\frac{n+3-L}{2^{L+1}}$ when $L \lt n$ the expected number of runs of exactly length $n$ is $\frac{1}{2^{n-1}}$...


3

For $P(X_3=2),\;$ for instance there is only one possible pattern, and we can compute number of ways as [Choose baskets for placing] $\times\;$[Place balls in baskets] $3-3-1-0-0-0-0:\;\left[\binom72\binom51 \right] \times \frac{7!}{3!3!}=14,700$ You can similarly find out for $X_3 = 0,\;$and $\;X_3=1.\; X_3>2$ will obviously have zero ways Finally ...


3

Convergence of $X_n$ in $L^1$ does not imply convergence in $L^2$. For a counterexample, let $$\mathbb P\left(X_n = n^{\frac12}\right)= n^{-1} = 1-\mathbb P(X_n=0). $$ Then $$\mathbb E\left[|X_n|\right] = n^{-1}n^{\frac12} = n^{-\frac12}\stackrel{n\to\infty}\longrightarrow 0, $$ so that $X_n\stackrel{L^1}\longrightarrow 0$, but $$\mathbb E\left[\left|X_n\...


3

Well, there's not much leeway, so it's probably most efficient to count by hand: $8+8+4=20$, $8+7+5=20$, $8+6+6=20$, $7+7+6=20$ – that's it, $4$ ways.


3

No, there is no such random variable $\xi$ and constant $c \neq 0$ so that $\xi$ and $\xi + c$ are distributed the same. Why? If these two random variables were distributed the same, then their CDF's (say $F_1$ and $F_2$) would also be the same. Now $$ F_1(z) = P(\xi \leq z) = P(\xi + c \leq z + c) = F_2(z+c) = F_1(z+c). $$ Therefore, the CDF is periodic ...


3

Consider the following random variables: $$ P(X = -1) = P(X = 1) = 1/2\\ P(Y = 0 \mid X = -1) = 1 \\ P(Y = 1 \mid X = 1) = P(Y = -1 \mid X = 1) = 1/2 $$ Then $E(XY) = E(X)E(Y) = 0$.


3

Depends on what valid claim numbers are. If the digits are randomly generated then the odds of them putting that one down are one in $100$ million. If the claim number needs to start in $34-$ then it's one in a million. But, I think this is just a couple of nails in a big coffin for your case. Having your company's name on the document is much more ...


2

You are correct: since the marginal pdf $f_X(x)$ can be obtained from the joint pdf by $$ f_X(x)=\int_{\mathbb{R}}f(x,y)\;dy $$ it follows that $$ \int_{\mathbb{R}^2}xf(x,y)\;dxdy=\int_{\mathbb{R}}x\Big[\int_{\mathbb{R}}f(x,y)\;dy\Big]\;dx=\int_{\mathbb{R}}xf_X(x)\;dx=\mathbb{E}[X] $$


2

Assuming $\delta=1-\epsilon\ge0$, for $x\ge0$ we have \begin{align} \def\pro#1{\textsf{Pr}\left(#1\right)} \pro{X\ge x}&=\pro{R\cos\Theta\ge x}\\ &=\pro{\cos\Theta\ge\frac xR}\\ &=\int_\delta^1\pro{\cos\Theta\ge\frac xr}\mathrm dr\\ &=\begin{cases} \frac1{\epsilon\pi}\int_\delta^1\arccos\frac xr\mathrm dr&0\le x\le\delta\\ \frac1{\epsilon\...



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