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6

The intuition is that if $g(x) \geq h(x) ~\forall x \in \mathbb R$, then $E[g(X)] \geq E[h(X)]$ for any random variable $X$ (for which these expectations exist). This is what one would intuitively expect: since $g(X)$ is always at least as large as $h(X)$, the average value of $g(X)$ must be at least as large as the average value of $h(X)$. Now apply ...


5

I think it's pretty hard to find a book which covers martingale theory; usually, books either give just an introduction or they focus on one particular aspect of martingale theory. I'll list some books which might be of interest and sketch (roughly) which parts they cover: David Williams: Probability with Martingales (Basic properties, optional stopping, ...


4

There are various forms for the remainder term of a finite Taylor expansion. One of them is $$f(x)=\sum_{k=0}^nf^{(k)}(a){(x-a)^k\over k!} +\int_a^xf^{(n+1)}(t){(x-t)^n\over n!}\ dt\ ,\tag{1}$$ whereby $f$ is assumed sufficiently differentiable in the neighborhood of $x=a$. This is of course valid as well if $f$ is complex-valued. For the proof of $(1)$ ...


3

By the fundamental theorem of calculus, $$ e^{ix} = 1 + (ix) \int_{0}^{1} e^{iux}\, du, $$ which is the case $r = 1$ of $$ e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^{k}}{k!}\right] + \frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r-1} e^{iux}\, du. \tag*{$P(r)$} $$ (N.B. $(1 - u)^{r-1}$ in the integrand, not $(1 - u)^{r}$.) Assume inductively that $P(r)$ ...


3

An alternative intuition is to see it as 4 groups of 13 with 4 friends all in different groups. The first can be in any group, the 2nd has 39 permissible slots out of 51, and so on. Thus Pr = $\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$


3

The solution is arguably simpler if you distinguish between permutations within the hands. There are $4!$ ways of distributing the aces over the $4$ players, and $13^4$ possible positions for the aces in the hands. That leaves $48!$ possibilities for filling the remaining $48$ slots. The size of the sample space in this perspective is just $52!$, so the ...


3

Because most of the time we identify functions which are almost surely equal. The whole theory of Lebesgue spaces carries this identifications, and for good reasons; there are a lot of properties that hold; for example we have $X = Y \ a.s. \implies E(X) = E(Y)$ In general we don't care too much about sets of measure zero; many theorem only prove things ...


3

That "Wikipedia" definition is for the special case where the covariance matrix is $\sigma^2 I$. In that case they are equivalent. A measure $\mu$ on $\mathbb R^n$ has density $\rho$ with respect to Lebesgue measure $\lambda^n$ iff $\mu(A) = \int_A \rho(x) \; d\lambda^n(x)$ for all Lebesgue measurable sets $A$.


3

We consider $m_n=\max\limits_{j=1}^n|a_j|$, $s_n^2=\sum\limits_{j=1}^na_j^2$, $W$ some random variable distributed like every $W_j$, and we simplify the quantities involved in Lindeberg's condition as follows; $$\begin{align} \sum_{j=1}^nE( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \}) &= \sum_{j=1}^n \frac{a_j^2}{n} E\left( W^2 \cdot ...


3

There is a mistake in the computation of the characteristic function. By definition, $$\varphi_Z(t) = \mathbb{E}e^{\imath \, t Z}, \qquad t \in \mathbb{R}.$$ Therefore, $$\begin{align*} \varphi_{\frac{Y}{\sqrt{\lambda}}}(t) &= \mathbb{E} \exp \left( \imath \frac{t}{\sqrt{\lambda}} Y \right) = \exp \left( \lambda \left[ \exp \left(- ...


3

$F_Y(y) := P(Y\le y)$ is the probability that the random variable $Y$ is less than or equal to a given real value $y$. This function is then known as the cumulative distribution function. For a continuous random variable it is the integral of the probability density function up to $y$, while for a discrete random variable it is the partial sum up to $y$ ...


3

You define conditional probability by the Kolmogorov definition. From the Kolmogorov definition, you can prove $P(A\cap B) = P(A\mid B)P(B)$. Then you just have to define what you mean when you write $P(A,B)$. You've defined it to mean $P(A,B)=P(A\cap B)$. Thus, from the definition of conditional probability and the definition of $P(A,B)$, you can prove the ...


3

Hints: Since $(W_t)_{t \geq 0}$ is a Brownian motion, it has independent normally distributed increments; in particular, $W_3-W_2, W_2-W_1,W_1$ are independent random variables which are Gaussian with mean $0$ and variance $1$. This implies that $(W_1, W_2-W_1,W_3-W_2)$ is (jointly) Gaussian with mean $\mu=0$ and covariance matrix $$C = \begin{pmatrix} 1 ...


3

Using the martingale property, $$ \begin{align} \text{E}(W_n) &= \text{E}[\text{E}(W_n | W_{n - 1})] \\ &= \text{E}(W_{n - 1}) \\ &= ... \\ &= \text{E}(W_1) , \end{align} $$ and so you have a uniform bound on both $\text{Var}(W_n)$ and $\text{E}(W_n)^2$.


3

The paper "On The Bound of Proximity of the Binomial Distribution to the Normal One" - Nagaev, Chebotarev (2010) has improvements to $C$ specifically for the Binomial Distribution. Theorem 2 on page 3 gathers together the results in the paper to show that $C$ can be taken to be .4215 in general. The paper notes that an Esseen (1956) paper demonstrates that ...


2

For (b), we want $E(X^2-4XY+4Y^2)$, which is $E(X^2)-4E(XY)+4E(Y^2)$. However, we cannot say in general that $E(XY)=E(X)E(Y)$. To calculate $E(XY)$, recall perhaps that $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)$. We know the covariance, and $E(X)$ and $E(Y)$, so we know $E(XY)$. For (c), as you know the $\pi$ makes no difference. We use the formula ...


2

Suppose your probability space is $(\Omega, \mathscr{F}, P)$ and $X$ be a random variable defined on this probability space. Technically, $$P(X= x) = P(\{\omega \in \Omega: X(\omega) = x\}) := P(E) $$ Having said $X$ is a random variable, $E = X^{-1}((-\infty, x])\backslash X^{-1}((-\infty, x))\in \mathscr{F}$, hence it does make sense to say the ...


2

The statement is "The probability that the random variable $X$ takes the value $x$ is equal to $\theta$." The notation is a bit misleading. You are right that $X$ does not equal $x$, but rather, after a measurement of $X$ is made, the value $x$ is observed as a result.


2

The short answer is that it's against the rules of probability $$P(a) + P(b) = P(a\vee b) + P(a\wedge b)$$ It's a theorem end of story. That's maths, in real life it's more a question of what probability means and why you should believe the rules, which is philosophically quite deep. There's an easy answer though, if you believe the above then I can con ...


2

The point mass at $0$ contributes an impulse of area p to the pdf $f_X(x)$. $$f_X(x) = p\delta(x) + f(x)$$ where f(x) is the part of $f_X(x)$ on (0,a]. When you convolve g with that impulse, you get a copy of g back multiplied by p. That adds on to the convolution of the rest of f with g since convolution is linear. $$h(z) = \int_0^{\infty}[p\delta(x) + ...


2

See e.g. Cryptographically secure pseudorandom number generator. A high-quality pseudorandom generator will be extremely difficult to distinguish from random. However, in principle, if you are lucky enough to guess which algorithm your opponent is using (including the size of the seed), and have a lot of time on your hands, you could try all possible seeds, ...


2

You can solve it using Ito's lemma but I'll rewrite it in a slightly different way to avoid the issues you came upon. We want to show $dW(t) = t^2dZ(t)$. This is really shorthand for: $$W(t) = W(0) + \int_0^t s^2 dZ(s)$$ You are also given $$W(t) = t^2Z(t) - 2\int_0^t s dZ(s)$$ We can imply from there that $W(0) = 0$ and so our ultimate goal is to show ...


2

Rather than trying to rephrase, I think a great choice is the explanation by F. Scholz here:


2

Square integrable variables are not any random variables. They are in fact pretty regular ! Once you know that your variable has a variance, it's natural that the distance to the mean of your variable can be controlled in probability by this variance. Chebyshev's inequality is probably the simplest way to achieve that.


2

Defining events: $v_1$: a vanilla cupcake is selected at first $v_2$: a vanilla cupcake is selected at second $b_1$: a vanilla cupcake is selected from box 1 $b_2$: a vanilla cupcake is selected from box 2 It is asked about $P(v_2|v_1)=\frac{P(v_1 \cap v_2)}{P(v_2)}$ $P(v_1 \cap v_2)$: Probability, that two vanilla cupcakes are selected. It is ...


1

Let's define the following events: the event that first box is chosen is $B_1$, the event that the second box is chosen is $B_2$, the event that the first cake is vanilla is $V_1$, the event that the second cake (from the same box) is vanilla is $V_2$. What you want to find is exactly $\Pr(V_2 | V_1)$. First, we have, by definition, \begin{align} ...


1

You have realised that P(first vanilla came from box 2) = $\frac{2}{3}$, and that only the 2nd box can contribute to the indicated probability. "Now I am stuck." Just continue t0 $\frac{2}{3} \times$ P(draw another vanilla from box 2) = $\frac{2}{3}\times \frac{1}{3} = \frac{2}{9}$


1

Working under condition that the first cake chosen is vanilla: let $E$ denote the event that the box picked at random contains another vanillacake. There are $3$ vanillacakes that can have been chosen with equal probability and in $2$ of these cases the box contains another vanillacake (exactly one). This tells us that $P\left(E\right)=\frac{2}{3}$. If ...


1

The answer is $1/2^m$ or $0$, depending on whether $y'$ is in the range of $A$ or not. Try the case $m=1$ to get a feel for this.


1

You didn't specify a distribution; I'll assume uniform distributions throughout. $y'$ may or may not be in the image of $A$. If it is, then since $A$ has full column rank, there's exactly one $y\in\mathbb F_2^m$ such that $Ay=y'$, so the probability is $2^{-m}$. If it isn't, then the probability is $0$. If instead $y'$ is picked randomly, the probability ...



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