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## Hot answers tagged probability-theory

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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