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47

We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms, $$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$ by the definition of the complement. And by the Kolmogorov Axioms, we see that $$P(\emptyset) = ...


12

We have this sum: $$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$ First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then $$ \frac k {k!} = \frac 1 {(k-1)!} $$ so that $$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$ The two expressions ...


7

The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. $$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$


5

In discrete time at least, the definitions I'm familiar with are fairly straightforward. Given an increasing filtration $\{\mathcal{F}_n\}_{n=0}^{\infty}$, a process $\{X_n\}_{n=0}^{\infty}$ is adapted if each $X_n$ is $\mathcal{F}_n$-measurable. For predictable processes, the random variables are measurable with respect to slightly smaller ...


4

The second player has 8 choices for his first move, 6, for the second, etc., so in any game he has a total of $8 \times 6 \times 4 \times 2 = 384$ possible sequences of moves. If only one choice is correct at each point, which is an underestimation, then in 5 million games he would be expected to draw at least $\lambda = 5 \times 10^6 / 384 \approx 1.3 ...


4

This is easiest if we treat the people as being distinguishable. There are $7$ floors each person can go to, and since the floors are chosen independently, there are $7^5$ total ways for the people to get off of the elevator. There are $\binom{7}{5} = 21$ ways to choose different floors for all $5$ people, and having chosen these floors, there are $5!$ ...


4

Calling the persons $P_i$ there is a probability of $7^{-5}$ that person $P_i$ steps out on floor $i$ for every $i\in\{1,2,3,4,5\}$. There are $5!$ rearrangements for the persons that result in a step out of exactly one person at the floors $1,2,3,4,5$. Moreover there are $\binom75$ ways to choose $5$ floors. So we end up with probability ...


4

The answer is YES. Let me introduce some notations first. Suppose $X=(X_1,\cdots,X_n)$ is an $\mathbb{R}^n$-valued random variable (i.e., a random vector). The characteristic function for $X$, denotes as $\varphi_X(u)$, is a function from $\mathbb{R}^n$ to $\mathbb{R}$: $$\varphi_X(u):=E(e^{iu\cdot X}),\quad u\in\mathbb{R}^n$$ where $u\cdot X$ is the ...


3

Your axiom list states that Additivity: $P(E_1 \cup E_2)=P(E_1)+P(E_2)$, where $E_1$ and $E_2$ are mutually exclusive. Note that $A \wedge \neg A$ is mutually exlusive with itself. Hence $$P(A \wedge\neg A)=P\left((A \wedge \neg A)\cup(A \wedge \neg A)\right) = P(A \wedge\neg A) + P(A \wedge\neg A) $$


3

As booleans, boolean-valued functions, boolean-valued random variables, or any other similar sort of thing, $A \wedge \neg A$ is the same thing as "false", and $A \vee \neg A$ is the same thing as "true". Applying this, your question reduces to Why is $P(\text{true}) = 1$ and $P(\text{false}) = 0$? Answering this depends on the exact details you want ...


3

Your intuition is pointing you in the right direction. Let $\mu_Y$ be the distribution of $Y$. Then $$ \infty>E|X+Y|=\int_{\Bbb R}E|X+y|\,\mu_Y(dy), $$ which means that $E|X+y|<\infty$ for $\mu_Y$-a.e. $y\in\Bbb R$. You should be able to finish it from there.


2

Regarding the indicator functions, just use their definition, i.e. they are equal to 1 on the event they indicate, and $0$ on its complement. Then $$ 1_{\{S\ge K_1\}}1_{\{S\ge K_2\}}=1_{\{S\ge K_1\}\cap{\{S\ge K_2\}}}=1_{\{S\ge \max\{K_1,K_2\}\}}. $$


2

Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$. Now observe that \begin{align} \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\ &= \operatorname{E}( Y_1+Y_2+\cdots Y_n)\\ &=\operatorname{E}( ...


2

Consider $X = \frac{1}{2}$ and $X_{\max} > \frac{1}{2}$. Then $a \in (0,1]$ and so \begin{align*} aX \le X =\frac{1}{2} \implies 1 - aX \ge \frac{1}{2} >0. \end{align*} Thus the random variable $\frac{X}{1-aX}$ is strictly positive, and so is it expectation. For a more general answer, if $X >0$, then we have \begin{align*} \begin{cases}1 -aX \ge 1 ...


2

Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"


2

The domain of $f(X)$, or actually $f\circ X$ is $\Omega$. It must be looked at it as the function $\Omega\to\mathbb R$ prescribed by $$\omega\mapsto f(X(\omega))$$ Defined like that also $f(X)$ is a random variable if $f:\mathbb R\to\mathbb R$ is a Borel-measurable function.


2

Consider $([0,1],\mathscr F ($borel sets on [0,1]$), m($lebesgue measure$))$ Let $X_n(x)=n1_{[0,\frac{1}{n}]}(x)$. This R.V. converges to 0 a.s. (it only does not converge to 0 for x=0) but $E(X_n(x)) = 1$ for every $n$ so $E(X_n) \not \rightarrow E(X) =0 $ Let $X_1=1_{[0,\frac{1}{2}]}$, $X_2=1_{[\frac{1}{2},1]}$, $X_3=1_{[0,\frac{1}{3}]}$, ...


2

Note that the graph is distributed as $G(n, p')$ for some $p'$. Why? Clearly the appearance of an edge in the final graph is independent of the choices for the other edges. Fix an edge $e$ and so $p' = P(e \text{ appears in final graph})$. Now, note that $$ \{e \text{ appears in final graph}\} = \{e \in \text{intermediate graph}\} \cup \{e \notin ...


2

Method 1: Using the Law of Total Probability. Let $S$ be the event of selecting a child with a sister. Let $F_{\rm GGG}, F_{\rm GGB}, F_{\rm GBB}, F_{\rm BBB}$ be the event of selecting a child from the relevant family structure; given that we know we are selecting from families with three children. The count of girls in a family has a Binomial ...


2

Both methods are fine. You didn't really specify the problem, but I gather you meant to imply that a child is selected from the family uniformly randomly, and that the genders of the three children are independent. In that case, both your arguments are correct.


2

This is because both $X_n$ and $Y$ only take values from a set of two points $\{3, 8\}$, which means given any positive $\epsilon \in (0, 5]$, the set \begin{align} \{\omega: |X_n(\omega) - Y(\omega)| \geq \epsilon\} & = \{\omega: X_n(\omega) = 8, Y(\omega) = 3\} \cup \{\omega: X_n(\omega) = 3, Y(\omega) = 8\} \\ & = \{\omega: X_n(\omega) \neq ...


2

We could consider "the other kids" as follows. Think of the group that Marcelle is in. There are $\binom{59}{29}$ equally likely ways to choose her classmates. Now let us count the favourables, the ways in which her four friends are in her group. We need to choose $25$ people from the remaining $55$. This can be done in $\binom{55}{25}$ ways. Thus the ...


1

Obviously we need $\rho \le 1$. Take a random vector $\mathbf{X}\sim \mathcal{N}(0,I_{n\times n})$ and a real RV $Z\sim \mathcal{N}(0,1), Z \perp \mathbf{X}.$ Then it is easy to check that $\sqrt{\rho}\cdot Z \cdot \mathbf{1}_{n\times 1}+ \sqrt{1-\rho}\cdot \mathbf{X}$ has the distribution you want.


1

With the well deserved criticism of the problem in my Comments, let's see how one would solve it if we ignore some of the unnecessary clauses in the problem and we make it work by changing one of the inputs. Let's say only 60 customers buy cake at Cafo (to make the numbers work with the other conditions), and the problem doesn't say stupid things like "each ...


1

If you wish to construct product measure with infinitely many probability spaces as factors, then no topological assumptions are needed. More precisely, if $\{(E_t,\mathcal E_t,\Bbb P_t): t\in T\}$ is a non-empty collection of probability spaces indexed by some set $T$, then there is a unique probability measure $\Bbb P$ on the product space $(\times_{t\in ...


1

The problem is not to construct an infinite product of sigma-algebras. This always exists and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable. The problem is to find measures on this space. Of course a measure on the infinite product always pushes forward to a measure on the finite products contained in ...


1

What about $X_n = n$ a.s., and $S\colon \mathbb{R}\to\mathbb{R}_+$ defined by $$S(x) = \frac{x^2}{x^4+1}$$? You will get $S(X_n) \to 0$ a.s., but clearly not what you want for $(X_n)_n$. Yet $S\geq 0$ is continuous and cancels only at $0$. It looks like a sufficient condition for what you want is that $S$ be continuous and strictly monotone with ...


1

You were mostly okay, except for the cases where you have three ones or three fives.   (Also note the multinomial coefficient in the other cases .) $$\begin{equation*} \begin{split} &\sum_{i\in\{2,3,4,6\}} \tfrac{5!}{2! \cdot 2! \cdot 1!}~ (\tfrac{2}{10})^2~(\tfrac{2}{15})^2 ~\mathsf P(X{=}i)^1 ~+~\tfrac{5!}{3!\cdot ...


1

Since $W\leq Z$, if $W>\varepsilon$ then $Z>\varepsilon$. In other words $$ \{W>\varepsilon\}\subset \{Z>\varepsilon\}$$ hence $\mathbb{P}(W>\varepsilon)\leq \mathbb{P}(Z>\varepsilon)$.


1

As Wikipedia points out, there are several results called Riesz (representation) theorem. A common feature of some of them is that on some spaces, every linear functional is integration against something. The result that the author uses here is the representation of linear functionals on $L^q$: they are integrals against $L^p$ functions. More precisely, ...



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