# Tag Info

8

Note that the integral can be written as $$\int_0^\infty \int_x^\infty f(t)\,\mathrm dt\,\mathrm dx=\int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dt\,\mathrm dx$$ and using Fubini we get $$\int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dx\,\mathrm dt=\int_0^\infty \int_0^t f(t)\,\mathrm dx\,\mathrm dt.$$ Here ...

5

To summarize: With replacement (the question), $(\alpha,\beta)$ is independent and $(x_\alpha,x_\beta)$ is not. Without replacement, $(\alpha,\beta)$ is not independent and $(x_\alpha,x_\beta)$ is. To show the case with replacement, consider $$P(x_\alpha\in A,x_\beta\in A)=P(x_\alpha\in A,x_\beta\in A,\alpha\ne\beta)+P(x_\alpha\in A,\alpha=\beta),$$ ...

5

Since the $X_i$ are independent, we have $E[X_i X_j X_k X_l] = E[X_i] E[X_j] E[X_k] E[X_l]$. But all four factors on the right side equal $\mu$, which is 0. Likewise, $E[X_i X_j X_k^2] = E[X_i] E[X_j] E[X_k^2]$. But $E[X_i]$ and $E[X_j]$ are both 0. Finally, $E[X_i X_j^3] = E[X_i] E[X_j^3]$. But $E[X_i] = 0$. So it isn't that there's some inherent ...

4

There must be at least three games for somebody to win, and cannot be more than five as somebody would have already won. One approach (not the most efficient in general) would be to list the possible outcomes AAA AABA AABBA AABBB ABAA ABABA ABABB ABBAA ABBAB ABBB BAAA BAABA BAABB BABAA BABAB BABB BBAAA BBAAB BBAB BBB Those of length $3$ each have a ...

4

As mentioned by @Nate, the key notion here is called coupling. Simply put, this means that there exists two random objects $G_p$ and $G_q$ defined on a common probability space $(\Omega,\mathcal F,\Pr)$ such that: $G_p$ is distributed as $A_p$ $G_q$ is distributed as $A_q$ $G_p\subseteq G_q$ with full probability Assuming these random objects exist, one ...

4

$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$

4

Let $p_t$ denote the PDF of $B(t)$ and assume that $U$ is independent of $B$ with PDF $f_U$, then the distribution of $B(U)$ has PDF $$q(\ )=\int p_t(\ )f_U(t)\mathrm dt.$$ In the present case, $U$ is uniform on $(0,1)$ and, for every $t$ in $(0,1)$, $$p_t(x)=\frac1{\sqrt{2\pi t(1-t)}}\mathrm e^{-x^2/(2t(1-t))},$$ hence $$q(x)=\int_0^1\frac1{\sqrt{2\pi ... 4 If \varphi(y)=E[X\mid Y=y\,] is the non-random function of y, then E[X\mid Y\,] is defined to be the random variable \varphi(Y). 4 You are asked to prove that if X\in L^q then there exists some finite C such that, for every \lambda\gt0, P(|X|\geqslant\lambda)\leqslant C\lambda^{-q}. And that if this latter condition holds then X\in L^p. Edit: It appears now that, in contradiction to the actual text of the question, the OP's problem is not to understand the meaning of these ... 4 Part (a) can only be solved when the probability space has an atom, that is, when there exists some measurable K\subseteq\Omega such that P(K)\gt0 and, for every measurable A\subseteq K, P(A) is either 0 or P(K). Consider$$X_n=n\cdot\mathbf 1_K,$$and let \delta=P(K). Then, for every event A, if P(A)\lt\delta then P(A\cap K)=0 hence ... 3 Here is a more complicated way to get to the result. This will broaden your knowledge: Assume X is the possibility that player A wins. His favorable rolls are 5 (15,24,33,42,51) out of the 36 possible, 6 rolls will make him lose, while the rest 25 will make the game a tie so far and they will roll again. On the next roll the probability will be the same. ... 3 For simplicity, assume the probability measure can be represented by a density p(u). Then we want to find g(x) to solve: f(r) = \int_{-\infty}^{\infty} g(\alpha r + u)p(u)du (for all r \in \mathbb{R}) This reminds me of a convolution. So define: y(x)=g(-x), with Laplace transform Y(s)=\int_{-\infty}^{\infty} y(x)e^{-sx}dx = G(-s). ... 3 Here is the step I cannot prove for you: The best way of maximising the probability of winning is always to bet the maximum amount of money which is smaller than your goal. The reason behind is that the odds is against you so you should try to win as quickly as possible by making fewest bets possible. The intuition is the more money you bet, the more the ... 3 An answer depends also on your ultimate objective as well as your bet sizing and they are dependent. The objective could be to play until you have doubled your money, play until your adversary is bankrupt, etc. For comparison, if you bet one 1 unit each round you can use the Gambler's Ruin solution where M_1+M_2 is your target:$$P_{ruin} = ...

3

Here is one way of doing it, but you need to be able to justify an interchange of an iterated integral. You have $F(x) = \int_0^x f(x) dx$ and $EX = \int_0^\infty x f(x) dx$. Write $x = \int_0^x dt$ to get $EX = \int_{x=0}^\infty \int_{t=0}^x f(x) dtdx$. All the quantities are positive, so we can switch the order of integration to get $EX = ... 3 Yes, the statement is true if the random variables are non-negative. Hint: By the triangle inequality and the positivity of the random variables, we have $$(X_n+X)-|X_n-X| \geq 0.$$ Now write $$2 \int X \, d\mathbb{P} = \int \liminf_{n \to \infty} (X_n+X-|X_n-X|) \, d\mathbb{P}$$ and apply Fatous lemma. 3 The core reason is that $$\color{red}{\sup_\color{black}{n}}\color{green}{\Pr}(|X_n-X|\geqslant\varepsilon)\qquad\text{and}\qquad \color{green}{\Pr}(\color{red}{\sup_\color{black}{n}}|X_n-X|\geqslant\varepsilon)$$ have little in common. In general, the latter is (much) larger than the former. For instance, if$\Pr(X_n=1)=1-\Pr(X_n=0)=1/n$for every ... 3 Write $$n \, 1_{(X_1^2>\epsilon^2n)}\leq {X_1^2\over\epsilon^2} \, 1_{(X_1^2>\epsilon^2n)}.$$ Now integrate both sides, and use dominated convergence to conclude$n\mathbb{P}(|X_1|>\epsilon \sqrt{n})\to0.$The sequence of random variables${X_1^2\over\epsilon^2} \, 1_{(X_1^2>\epsilon^2n)}$converges to zero pointwise as$n\to\infty$. Since ... 3$\textbf{My Question is:}$(...) are$Y\mathbf1_{|Y|\le a}$and$-Y\mathbf1_{|Y|>a }$independent ? Of course not. Let$U=Y\mathbf1_{|Y|\le a}$and$V=-Y\mathbf1_{|Y|>a }$, then$[U=0]=[|Y|\gt a]\cup[Y=0]$,$[V=0]=[|Y|\leqslant a]$, and$[U=V=0]=[Y=0]$. What does all this tell you about the possibility that$(U,V)$is independent? An extended ... 3 Let$hbe an arbitrary bounded Borel function then \eqalign{ \mathbb{E}(h(Z))&=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \underbrace{\frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(-y)e^{-y^2/2}dy}_{y\leftarrow-y}\cr &=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(y)e^{-y^2/2}dy\cr ... 3 Let X_{i} be i.i.d. and Y_{n} = \sum_{i=1}^{n} X_{i}. It is well-known that If \Bbb{E} X exists in [-\infty, \infty], then S_{n}/n \to \Bbb{E} X a.s. If \Bbb{E}|X| = \infty, then in probability 1, S_{n}/n does not converge in (-\infty, \infty). (The first case is SLLN when \Bbb{E}|X| < \infty, and the case \Bbb{E}X = \pm\infty are ... 3 Yes. You know that the set B+c is the inverse image of the measurable set, B, under the continuous function: f(x)=x-c, and since continuous functions are measurable, and the inverse image of a measurable set is measurable, B+c is also measurable. Note: there is nothing particularly special about Lebesgue measure here other than it being a Haar ... 3 If (A_n) is a martingale with respect to (B_n), then, in particular, each random variable A_n is measurable with respect to the sigma-algebra \mathcal B_n=\sigma(B_i;i\leqslant n) since A_n=E(A_{n+1}\mid\mathcal B_n), hence \mathcal A_n=\sigma(A_i;i\leqslant n) is such that \mathcal A_n\subseteq\mathcal B_n. Then the hypothesis that ... 3 The A_k are independent because for any i \neq j, the sets of tosses 2^i, \ldots, 2^{i+1}-1 and 2^j, \ldots, 2^{j+1}-1 are disjoint. Hence, A_i tells you no information about A_j. Let's find upper and lower bounds for P(A_k). Amongst the tosses numbered 2^k, \ldots, 2^{k+1}-1, there are 2^k tosses, and so, there are 2^k-k+1 blocks of ... 3 Your definition of \tau is flawed, actually,\tau=\int_0^\infty\mathbf 1_{W_t\gt t}\,\mathrm dt,$$hence$$E(\tau)=\int_0^\infty P(W_t\gt t)\,\mathrm dt.$$For every t, W_t is normal centered with variance t hence$$P(W_t\gt t)=P(Z\gt\sqrt{t}),$$where Z is standard normal. By symmetry, P(Z\gt\sqrt{t})=P(Z\lt-\sqrt{t}) hence$$ ... 3 The functiont\mapsto\frac1{1+t}$is convex on$t\geqslant0$hence $$E\left(\frac1{1+X^2}\right)\geqslant\frac1{1+E(X^2)}=\frac1{1+\nu}.$$ The lower bound is attained when$P(X=\sqrt\nu)=P(X=-\sqrt\nu)=\frac12$. On the other hand, if$P(X=0)=1-\frac\nu{x^2}$and$P(X=x)=P(X=-x)=\frac\nu{2x^2}$for some$|x|\geqslant\sqrt\nu$, then$E(X)=0$and ... 2 Total variation norm is a norm on measures whereas all the other types of convergence/norms/topologies that you mentioned are on spaces of functions. Nonetheless functions can be measures too. If you have a sequence of measures$\mu_n$and say all of them are absolutely continuous with respect to a third measure$\lambda$, then the measures converge in TV ... 2 If$(X_n)_{n\in\mathbb N}$is a sequence of random variables with nondegenerate normal distributions, then$P(X_n\in\mathbb Z)=0$for every$n$, hence $$P(\exists n\in\mathbb N,\,X_n\in\mathbb Z)\leqslant\sum_{n=1}^\infty P(X_n\in\mathbb Z)=0,$$ that is, $$P(\exists n\in\mathbb N,\,X_n\in\mathbb Z)=0.$$ This does not assume independence, only that the ... 2 Note that $$(1-p_n)^{\lfloor x/p_n \rfloor} = (1-p_n)^{x/p_n} (1-p_n)^{\lfloor x/p_n \rfloor - x/p_n}$$ For the first factor, we use the fact that$(1-\epsilon)^{x/\epsilon} \to e^{-x}$as$\epsilon \to 0$. This is a standard fact from calculus and there are many ways to check it. Note that we can write$(1-\epsilon)^{x/\epsilon} = ...

2

Here is a more elegant answer than Henry's, but you need more work. Let $n$ be the number of wins that one player needs to obtain for the game to end. Start at the single "1" at the apex of Pascal's triangle. For every game, go down one row, left if A wins and right if B wins. Notice that as long as each of A and B have less than $n$ wins, the number of ...

Only top voted, non community-wiki answers of a minimum length are eligible