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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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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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