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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 ...

4

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 = $\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$

4

Probability is not just defined on a set. It's also defined on a $\sigma$-algebra of subsets. Namely, a probability structure is made of three parts: The underlying set, say $X$. The sets to which you can assign probability. The function which assigns probability. There are several axioms which we require of these structures, such as the sets to which ...

4

If you wonder why so far only the "trivial" example of a Dirac measure has been mentioned, there is actually a good reason for this. The young Ulam wondered whether there is an uncountable set $X$ for which there is a probability measure $$\mu \colon \mathcal P(X) \rightarrow [0,1]$$ that vanishes on singletons (i.e. $\mu(\{x\})= 0$ for all $x \in X$). He ...

4

If $X,Y,Z$ are independent then: $$\mathbb{P}[X \in A, Y \in B Z \in C] =\mathbb{P}[X \in A] \mathbb{P}[Y \in B] \mathbb{P}[Z \in C]$$ consider now $s^1(x) = \sum_{i =1}^{k_1} 1_{A_i}(x) \alpha_i$ $s^2(y) = \sum_{j =1}^{k_2} 1_{B_j}(y) \beta_j$ $s^3(z) = \sum_{l =1}^{k_3} 1_{C_l}(z) \gamma_j$ Now compute $$\mathbb{E}[s^1(X) s^2(Y) s^3(Z)] = ... 4 We have to consider three cases separately: S_{k-1}(\omega)<0: Then S_k(\omega) \leq 0 and therefore$$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$S_{k-1}(\omega)=0: Then |S_k(\omega)|=1 and therefore$$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$S_{k-1}(\omega)>0: Then S_k(\omega) \geq 0 and therefore ... 3 Using the martingale property,$$ \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} $$and so you have a uniform bound on both \text{Var}(W_n) and \text{E}(W_n)^2. 3 It is true that the second property can be deduced from the first one. Indeed, the first property implies that (B_t) has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction: Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary ... 3 That "Wikipedia" definition is for the special case where the covariance matrix is \sigma^2 I. In that case they are equivalent. A measure \mu on \mathbb R^n has density \rho with respect to Lebesgue measure \lambda^n iff \mu(A) = \int_A \rho(x) \; d\lambda^n(x) for all Lebesgue measurable sets A. 3 For any cardinal \kappa, you can define the product measure on 2^{\kappa} where 0, 1 are assigned one half measure each. This measure is defined on the sigma algebra generated by clopen subsets of 2^{\kappa}. Let us call the corresponding measure algebra \text{Random}(\kappa). The class of measure algebras thus obtained forms the building block for ... 3 The solution is arguably simpler if you distinguish between permutations within the hands. There are 4! ways of distributing the aces over the 4 players, and 13^4 possible positions for the aces in the hands. That leaves 48! possibilities for filling the remaining 48 slots. The size of the sample space in this perspective is just 52!, so the ... 3$$F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(X\leq x)F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(Y\leq y)F_{X,Y}(x,y)^2\leq P(X\leq x) P(Y\leq y)$$2 Yes, otherwise how do you plan on taking intersection of sets which don't live in the same space? You should think of random variable as just measurable functions, which then would be kinda odd if they didn't share the same domain. Perhaps a better question is how do we know that, given a sequence of random variable, that there is a probability space which ... 2 Rather than trying to rephrase, I think a great choice is the explanation by F. Scholz here: 2 Let 0 represent one orientation and 1 represent the other orientation. Each arrangement of the n magnets then corresponds to an n-bit binary string. Two adjacent magnets repel if they are in opposite orientations and attract if they are in the same orientation. Thus, a block of magnets corresponds exactly to a block of identical bits. For n=8, for ... 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 Let us assume that \Omega is finite or countable. I think the main confusion here is the relationship between a probability measure and a random variable. A probability measure \mathbb P is a function \mathbb P:\mathcal F \rightarrow [0, 1] for some \sigma-algebra. A random variable, on the other hand, is a function X: \Omega \rightarrow S where ... 2 Consider x and y as the axes on the graph. Then all values of (x,y) will lie on the square [0,1] \times [0,1]. Now for any fixed value of x \in [0, 1], try to find the region of the square that has values of y such that |y - x | < 1/2. Observe that this region is a band around the line y = x and using simple geometry you can calculate its ... 2 The random variable \omega\mapsto \mathbb{I}_B(\omega)X(\omega) is not necessarily of the form g\circ X, so your formula for E(g\circ X) is not useful here. 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

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

$X,Y,Z\sim_{iid}\text{Exp}(1)$ because $F_{X,Y,Z}(x,y,z)=F_X(x)F_Y(y)F_Z(z)$ for all $(x,y,z)\in\mathbb{R}^3$. Denote $S_n=\sum_{i=1}^nX_i$ where $X_i\sim_{iid}\text{Exp}(\lambda)$. Distribution of $S_n/n$ follows $\text{Gamma}(n,n\lambda)$ because ...

2

$\lim_{n \to \infty} X_n$ is the pointwise limit of $X_n$. Actually the pointwise limit of $X_n$ is $0$ almost surely, it is $0$ everywhere except at $0$. Why? Except at $x=0$, $\lim_{n \to \infty} X_n(x) = 0$, since there always exists $n$ such that $\frac{1}{n} < x$. The expectation of a function that is $0$ almost surely is $0$.

2

Note that $$n^{\alpha} \mathbb{P}(|X|>n) = \mathbb{E}(n^{\alpha} 1_{\{|X|>n\}}) \leq \mathbb{E}(|X|^{\alpha} 1_{\{|X|>n\}}).$$ Now you can either apply the dominated convergence theorem or use the fact that $$\mathbb{E}(|X|^{\alpha} 1_{\{|X|>n\}}) \leq \sum_{k \geq n^{\alpha}} \mathbb{P}(|X| \geq k^{1/\alpha})$$ and the previous theorem.

2

A closed form expression is provided in the following paper. SV Amari, RB Misra, Closed-form expressions for distribution of sum of exponential random variables, IEEE Transactions on Reliability, 46 (4), 519-522.

2

You can do them more simply as: $P(AB)\leq P(A)=0$ hence $P(AB)=0$. But $P(A)P(B)=0$ so $P(AB)=P(A)P(B)$ implying $A,B$ are independent. $P(A\cup B)\geq P(A)=1$ hence $P(A\cup B)=1$. So $P(A)+P(B)-P(AB)=1\implies P(B)=P(AB)\implies P(A)P(B)=P(AB)$ so $A,B$ are independent. Suppose $D,D^c$ are independent. Then $P(DD^c)=P(D)P(D^c)\implies ... 2 Six ways to get$1,3,6\$: $$\begin{array}{c|c|c|c|c|c} \overbrace{ \begin{array}{rc} \text{first die:} & 1 \\ \text{second die:} & 3 \\ \text{third die:} & 6 \end{array}} & \overbrace{ \begin{array}{rc} \text{first die:} & 1 \\ \text{second die:} & 6 \\ \text{third die:} & 3 \end{array}} & \overbrace{ \begin{array}{rc} ... 1$$\mathbb{E}X=\mathbb{E}X^+-\mathbb{E}X^-=\mathbb{E}[X\vee 0]-\mathbb{E}[-X\vee 0]=\int_0^\infty\mathbb{P}\{X>x\}dx-\int_{0}^\infty\mathbb{P}\{-X\ge x\}dx$$1 Any open set U\subset\mathbb R^2 is a countable union of open rectangles (a,b)\times(c,d). Not every closed set C\subset\mathbb R^2 is a countable union of closed rectangles [a,b]\times[c,d]. The same thing happens with open and closed balls, too. In an open set you always have a little bit of room to find a rectangle around any point, but in a ... 1 We show directly that the two definitions are equivalent: We need only to show that all open sets in \mathbb R^n can be written as a countable union of open rectangles. Let V \in \mathbb R^n be open. Let C be its complement. Let Q be a countable dense subset in V. For each p \in Q, define$$r_p = \sup \{r >0 : (p_1-r, p_1+r) \times \cdots ...

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