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If $h:\mathbb R\to\mathbb R$ is Borel-measurable, then $h(Y)$ is measurable with respect to the $\sigma$-algebra generated by $Y$: $$\sigma(Y) = \{ Y^{-1}(B):B\in\mathcal B(\mathbb R)\}.$$ It follows from the definition of conditional expectation that $$\mathbb E[X h(Y)\mid Y] = h(Y)\mathbb E[X\mid Y]$$ with probability one.

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Recall that if $U\sim\mathsf U(0,1)$ and the distribution function $F_Y$ of $Y$ is absolutely continuous, then the random variable $F_Y^{-1}(U)$ has the same distribution as $Y$. In particular, if $F_Y(y) = (1-e^{-y})\mathsf 1_{(0,\infty)}(y)$ then $$F(Y) = U \iff Y = -\log(1-U).$$ Set $Y_k = -\log(1-X_k)-1$, then $Y_k\stackrel d= Y-1$ where $Y\sim\... -1 If$X_i\equiv\mathcal{U}(0,1)\to Y_i=1-X_i\equiv\mathcal{U}(0,1)$. Now, note that$\mathbb{P}(\{-\sum \ln Y_i\geq n\})= \mathbb{P}(\{- \ln \prod Y_i\geq n\})= \mathbb{P}(\{ \prod Y_i\geq e^{-n}\})You see the way to finish? 2 The first question is: Under what circumstances is $$\frac \partial {\partial\theta} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \bullet\bullet\bullet$$ the same as $$\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac \partial {\partial\theta} \bullet\bullet\bullet \text{ ?}$$ The next question is: Why is $$\frac \partial {\partial\theta} \... 3 Let us rewrite the product rule as follows:$$(fg)'=f'g+g'f=\frac{f'}{f}fg+\frac{g'}{g}fg=\left(\frac{f'}{f}+\frac{g'}{g}\right)fg$$Yours is just the generalization to n factors, but is handled in the exact same way. 0 No, it does not. Consider for example a Gaussian variable X\sim N(0, 1) f_X(x) = e^{-1/2\ x^2} \ne e^{it EX} = e^{0} = 1 BS 2 It is not true in general that \mathbb{E}[e^{itX}]=e^{it\mathbb{E}[X]}. For instance, suppose that X=1 with probability \frac{1}{2} and X=-1 with probability \frac{1}{2}. Then \mathbb{E}[X]=0, hence e^{it\mathbb{E}[X]}=1. On the other hand,$$ \mathbb{E}[e^{itX}]=\frac{1}{2}\Big(e^{it}+e^{-it}\Big)=\cos(t) $$0 For simplicity, let T_1 be the shelflife of molecule 1 after it enters the system, at time 0, and T_2 be that of molecule 2 after it enters the system, at time \Delta t. Then the event of molecule 1 decaying before molecule 2 is represented by: \{T_1< T_2+\Delta t\} Noting that the supports of the exponential distribution asserts that ... 1 The two word answer is "polar coordinates". In more detail, let f:S^{n-1}\to\Bbb R be a continuous function. Then$$ \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ... 1 Here is another way of thinking about it: condition on the location of the 1st Head and use the law of total expectation (tower rule). LetN$be the number of throws for getting the first HT. Let$Z$be the location of the first H. Then, by the tower rule: $$E[N] = E[E[N|X]] = \Sigma_{k = 1}^{\infty} E[N | Z = k] P(Z = k)$$. Let us compute each of ... 1 The other inclusion is in general false. If$X = 0$is constant, then$\sigma(X, Y) = \sigma(Y)$, while$\sigma(X, XY) = \{\emptyset, \Omega\}$. These are different for any nonconstant random variable$Y$. 0 I think it is better to avoid thinking in terms of bona fide white noise, as physicists often think. You should instead think in terms of computing $$I_l^m=\int_0^2 P_l^m(x-1) dW(x).$$ This is a stochastic integral. (I shift everything to start at zero for consistency with standard notation in math.) Approach 1: Rewrite the stochastic integral by "formal ... 1 The expected value of a normal distribution is the mean, in your case$\mu$, because the normal is symmetrical from let to right. For any probability density function$f(x)$this can be calculated by integrating$x\times f(x)$across its domain. The normal is defined as having the pdf: $$f(x)=\frac{1}{\sqrt{2\pi}}\exp{(\frac{-x^2}{2})}$$ The mean of that ... 1$\newcommand{\E}{\mathbb{E}}$$\newcommand{\P}{\mathbb{P}}For positive values, \psi(x)=x^{\alpha} is monotone (I think even for all \alpha \in \mathbb{R}, but certainly for 0 < \alpha < 1). Therefore for such \psi we have$$X > Y \text{ almost surely } \implies \E[\psi(X)]) > \E[\psi(Y)] \quad \text{and} \quad\E[X] > \E[Y]$$by ... 3 We can first solve the easier problem of finding the probability p_k that k uniformly random non-zero elements of \mathbb Z_q add up to 0. This probability satisfies the recurrence$$ p_{k+1}=\frac1{q-1}(1-p_k) $$with the initial value p_0=1, with solution$$ p_k=\frac{1-(1-q)^{1-k}}q\;. $$You're adding k non-zero elements with probability ... 0 By Linearity of expectation, we have$$\mathbb{E}[4X^2+4X+1] = 4\mathbb{E}[X^2]+4\mathbb{E}[X]+1.$$If you plug in \mathbb{E}[X] = 5.5, \mathbb{E}[X^2] = 46.5, and \mathbb{E}[4X^2+4X+1] = 61, the left side of the above equation is 61, but the right side is 4 \cdot 46.5 + 4 \cdot 5.5 + 1 = 209. So you have a mistake somewhere. I'll go ahead and ... 0 The first and second look good, but the third can't be right. Since expectation is linear in its argument (E(aX + b) = aE(X) + b),$$ E(4X^2 + 4X + 1) = 4E(X^2) + 4E(X) + 1 = 4(46.5) + 4(5.5) + 1 = 209. $$(You should double check this by calculating the third expectation directly again.) 0 You took the derivative wrong. Notice:$$\frac{dF_U(u)}{du}=\frac{d}{du}(-e^{\frac{-u}{\beta}})-\frac{d}{du}(\frac u\beta e^{\frac{-u}{\beta}})+\frac{d}{du}(1)=\frac 1\beta e^{\frac{-u}{\beta}}-\frac {1}{\beta}e^{\frac{-u}{\beta}}- \frac u{\beta}\frac {1}{(-\beta)} e^{\frac{-u}{\beta}}+0=\frac{1}{\beta^2}u e^{\frac{-u}{\beta}}$$which is the expected ... 3 I think you are quite over-complicating it. If X,Y,Z are uniformly distributed over [0,1] and independent,$$\begin{eqnarray*}E=\mathbb{E}[(X+Y+Z)^2]&=&\iiint_{[0,1]^3}(x+y+z)^2\,d\mu\\&=&\iiint_{[0,1]^3}(x^2+y^2+z^2+2xy+2xz+2yz)\,d\mu\\&=&3\cdot\frac{1}{3}+3\cdot 2\cdot \frac{1}{4}=\color{red}{\frac{5}{2}}.\end{eqnarray*}$$... 0 You know that \mathsf E(X)=\lambda^{-1} = 1/5 and \mathsf E(Y\mid X)=(X+3)/2 , from what was given. So you may make good use of the tower rule (aka the Law of Iterated Expectation).$$\mathsf E(Y)=\mathsf E(\,\mathsf E(Y\mid X)\,)$$1 The (conditional) expectation of Y is \frac{X+3}{2}. So you want$$\int_0^\infty \frac{x+3}{2}\cdot 5e^{-5x}\,dx.$$Here I am assuming that by the exponential with parameter \lambda you mean that the density function is \lambda e^{-\lambda x}, x\gt 0. More infrequently, some people think of it as having density \frac{1}{\lambda}e^{-x/\lambda}. 1 First, I provide a way to generate a uniform random vector on S^{n-1} and it will be used in our proof later. Fact: Let X_1, X_2, \cdots, X_n \sim \mathcal{N}(0, 1) and be independent. The the vector$$ X = (\frac{X_1}{Z}, \frac{X_2}{Z}, \cdots, \frac{X_n}{Z}) $$is a uniform random vector on S^{n-1} where Z = \sqrt{X_1^2 + \cdots + X_n^... 3 Assuming that U_i > 0, we may define X_i as \log U_i. Assuming that \mathbb{E}[\log U_i]<+\infty, the law of large numbers gives$$ \frac{X_1+\ldots+X_n}{n}\to \mathbb{E}[\log U_i] $$and by exponentiating back$$ V_n^{1/n} \to \exp\left(\mathbb{E}[\log U_i]\right). $$0 If X \ge Y, then$$E|X-Y| = EX - EY$$Otherwise$$E|X-Y| = EY - EX$$Neither of which are necessarily equal to$$E|X| - E|Y|$$1 3) Consider -\log(V^{1/n}_n) and use the fact 1 and 7 in : https://en.wikipedia.org/wiki/Exponential_distribution#Related_distributions 4) X_n converges in law to X means \forall g continuous bounded, E(g(X_n)) \to E(g(X)). Now, if you pick a continuous function f, what can you say about E(g(f(X_n))) ? 2 No, the required integral is$$\begin{align}\mathsf P(-2\ln(X)\leq t) =&~ \mathsf P(X\geq\mathsf e^{-t/2})\\[1ex] =&~ \mathbf 1_{\exp(-t/2)\in[0;1]}\int_{\exp(-t/2)}^1\operatorname d x \\[1ex]=&~ (1-\mathsf e^{-t/2})~\mathbf 1_{t\in[0;\infty)}\end{align}$$1 Using the comment given by Hagen von Eitzen, we get:$$ \int_{\mathbb{S}^{n-1}} \sum_{i=1}^n (\sigma_ix_i)^2 = \sum_{i=1}^n \int_{\mathbb{S}^{n-1}} (\sigma_ix_i)^2= \sum_{i=1}^n \sigma_i^2\int_{\mathbb{S}^{n-1}} x_i^2$$By symmetry, we see that \int_{\mathbb{S}^{n-1}} x_i^2 is independent of the index i, hence:$$\int_{\mathbb{S}^{n-1}} x_i^2=\frac{1}...

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The PDF of $T$ is $$P(T)=P(X)\left|\frac{dX}{dT}\right|$$ $$=1\times\left|\frac{dX}{dT}\right|$$ $$=\left|\frac{X}{2}\right|$$ $$=\frac{1}{2}e^{-T/2}$$ The CDF is $$P(T\le t)=\int_{0}^{t}\frac{1}{2}e^{-T/2}dT$$ $$=-\left[e^{-T/2}\right]_{0}^t$$ $$=1-e^{-t/2}$$

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You could find the joint distribution of $X$ and $Y$, and use that for the computation. However, it is less work to note that $|X-Y|=|2X-6|$. For each of the $7$ possible values $k$ of $X$, find $\Pr(X=k)$. Then our expectation is $$\sum_{k=0}^6 |2k-6|\Pr(X=k).$$ We start. For $k=0$, the probability is $\frac{1}{2^6}$, and $|2k-6|=6$. That gives a ...

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In order to apply dominated convergence, write the expectations as integrals over the probability space $(\Omega,{\cal F},P)$, not the real line: $$E(f(X_n))=\int_\Omega f(X_n(\omega))\,P(d\omega)\to\int_\Omega f(X(\omega))\,P(d\omega)=E(f(X)).$$

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The cumulative distribution function for $Y$ is: $$F_Y(y)=\operatorname{Prob}(Y<y)=\operatorname{Prob}(-\sqrt{y}<z<\sqrt{y})=F_Z(\sqrt{y})-F_Z(-\sqrt{y})=2F_Z(\sqrt{y})-1$$ Now to get the density differentiate with respect to $y$, and of course to differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ ...

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As the comment by Danielsen points out, the most common definition of the convergence in distribution is $$\lim_{n \rightarrow +\infty }\mathbb E\left[f\left(X_n\right)\right]=\mathbb E\left[f(X)\right]$$ for each continuous and bounded function $f$. If $f$ is such a function and $X_n\to X$ almost surely, then $X_n(\omega)\to X(\omega)$ for each $\omega\in\... 0 Copying John Dawkins' comment: 2 No. Let$(X_n,Y_n)$be$(1,1)$with probability$\frac{n-1}n$and$(n+1,1)$with probability$\frac1n$. Then$\frac{X_n}{Y_n}\to1$in probability, but $$\frac{\mathbb E[X_n]}{\mathbb E[Y_n]}=2\;.$$ I don't know what conditions would lead to$\frac{\mathbb{E}[X_n]}{\mathbb{E}[Y_n]} \rightarrow 1$. 0 Something that is closely related can be a statement like that: Claim: Let$X$and$Y$equivalent in the sense already described. If$U \in A_X$, than there exists$U' \in A_Y$such that the symmetric difference has zero measure:$P(U \Delta U')=0$. Trial proof: We first prove the statement for all sets of the type$U=\{X^{-1}(B)\}$where$B$is a Borel ... 0 One property of a discrete random variable is that the probabilies add up to one. In your case $$\sum_{x=1}^3 P(X=x)=1$$$P(X=1)+P(X=2)+P(X=3)=1\frac{2+5P}{5}+ \frac{1+3P}{5} + \frac{1.5+2P}{5}=12+5P+1+3P+1.5+2P=5$Solve for$P$. After you have evaluated the value for$P$you can calculate$P(X=x)$for every single value of$x$. Then the expected ... 2 Let$\left(y_k\right)_{k\geqslant 1}$be a decreasing sequence of continuity points of the cumulative distribution function of$X$, and convergent to$x$. Let$k\geqslant 1$be fixed. Then for$n$large enough we have$x_n\leqslant y_k$, hence$\mathbb P(X_n\leqslant x_n)\leqslant \mathbb P\left(X_n\leqslant y_k\right)$, and taking$\limsup_{n\to +\infty}$, ... 0 No. The moments only depend on the unknown parameters through$p$. Thus they don't contain any more information than does$p$. Fitting to the central moments is a complicated waste of time; all you can find out about the unknown parameters you can find out from$p$(or, if you prefer, from the second and third central moments, which together determine$p$). 2 For the first question, since$p+q=1$it follows that $$(p-q)^2=p^2-2pq+q^2=p(1-q)-2pq+q(1-p)=p+q-4pq=1-4pq$$ Therefore taking a square root yields$\sqrt{1-4pq}=|p-q|$. For the second question, while generating functions can be regarded as formal power series, they can also be viewed as defining an analytic function in some neighborhood of zero. There is ... 2 The notation you use,$X+X$, indicates that a single realization of$X$is added to itself to yield$2X$; thus this would represent the outcome of a single die roll, multiplied by$2$. If you wanted to represent the outcome of two independent die rolls, that is precisely described as $$X_1 + X_2,$$ where$X_1, X_2$are independent and identically ... 1$X+X=2X$everywhere. It is twice the value seen on 1 die. If you want the sum of values on 2 separate die,then$S=X+Y$;$X,Y\to iid \in\{1,\cdots 6\}$1 You should use$X+Y$, where$X$and$Y$are independent and identically distributed. Note that identically distributed does not mean identical. 1 For any integer$n\ge 1, consider the function \begin{align*} f_n(x) = \begin{cases} 1, & x \le \alpha-\frac{1}{n},\\ -n(x-\alpha), & \alpha-\frac{1}{n} < x \le \alpha,\\ 0, & x > \alpha. \end{cases} \end{align*} Thenf_n$is continuous and$f_n(x) \nearrow 1_{x \le \alpha}, and \begin{align*} E(f_n(X)) \rightarrow P(X \le \alpha). \end{... 1 You just need to note that \begin{align*} \{W_{12}\le t\} \cap \{W_{13} \le t\} &= \{X_1 \le t\}\cap\{X_2\le t\}\cap\{X_3 \le t\}. \end{align*} 0 To visualize the pdf of|X-Y|$you can draw a picture. After reading off the pdf it is not difficult to calculate the expected value. First the equation$|X-Y|\leq z$has to be solved. For this purpose two cases have to be regarded.$a) \ X-Y\geq 0$The absolute value signs can be dropped-without any manipulation. The inequality becomes$X-Y\leq z \...

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Let $A$ be a non-trivial subset of $\Omega$ and let $\Omega$ be equipped with $\sigma$-algebra $\left\{ \varnothing,A,A^{c},\Omega\right\}$ and probability measure $P$ determined by $P\left(A\right)=1$. Let $X:\Omega\to\mathbb{R}$ be prescribed by $\omega\mapsto1$ Let $Y:\Omega\to\mathbb{R}$ be prescribed by $\omega\mapsto1$ if $\omega\in A$ and $\omega\... 0 You want$\int_0^1\int_0^1 |x-y| d(x,y)=\int_0^1(\int_0^1(|x-y|dx)dy=\int_0^1(\frac{y^2+(1-y)^2}{2})dy=\int_0^1(y^2-y+\frac{1}{2})=\frac{1}{3}$0 Take a unit square in the$xy$-plane of 3-space. Draw the diagonal from$(0,0)$to$(1,1)$. Now above the point$(1,0)$draw a point at height$z =1$; above the point$(0, 1)$draw another point at height one. Connect each of these two points to the diagonal line you drew in the plane, forming two triangles. Those triangles are the graph of your function. ... 2 The squared variation of$X$from its mean could itself be considered a random variable, say defined by$Y=(X-\mu)^2$. The variation then measures the mean of$Y$, which involves using$Y$'s and hence$X$'s probability distribution. Your feelings wouldn't say to average a random variable (like$Y$) by simply adding the possible values it could take and ... 1 In any manner, if $$L(p)=\prod_{i=1}^n f_{X_i} (x_i; p)$$ then $$\log\left(L(p)\right)=\sum_{i=1}^n\log\left(f_{X_i} (x_i; p)\right)$$ $$\frac{L'(p)}{L(p)}=\sum_{i=1}^n \frac{f'_{X_i} (x_i; p)}{f_{X_i} (x_i; p)}$$ and since you want$L'(p)=0\$, the rhs seems (at least to me) simpler to manipulate.

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