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The approach using the ratio of densities is allright, provided one writes the joint density correctly. Considering $a=1/\theta$, you are told that $$f_{X,Y}(x,y)=2a^2\mathrm e^{-a(x+y)}\,\mathbf 1_{0\lt x\lt y},$$ hence $$f_X(x)=\int f_{X,Y}(x,y)\mathrm dy=\mathbf 1_{x\gt0}\int_x^\infty2a^2\mathrm e^{-a(x+y)}\,\mathrm dy,$$ that is, $$f_X(x)=2a\mathrm ... 2 If n is the number of rolls left before a given roll, after the roll, there are either n-1 or n+1 rolls left, with respective probabilities p=\frac56 and 1-p=\frac16. One asks for the mean number of rolls X_2 starting from 2 rolls left. For every n, the mean number of rolls X_n starting from n rolls left is n times the mean number of ... 2 We have Y_i\sim\mathcal{U}(\theta,\theta+1) and CDF of Y_i based on Wikipedia$$ G_{Y_i}(y)=\Pr[Y_i\le y]=\frac{y-\theta}{\theta+1-\theta}=y-\theta. $$Here, Y_{(n)} is n-th order statistics. Therefore, Y_{(n)}=\max[Y_1,\cdots, Y_n]. Note that Y_{(n)}\le y equivalence to Y_i\le y for i=1,2,\cdots,n. Hence, for \theta< y<\theta+1, the ... 2 You are asked \Pr\left[k<\dfrac{Y_{(n)}}{\theta}\le 1\right]. Now, take a look the part: k<\dfrac{Y_{(n)}}{\theta}\le 1. Multiply each side by \theta, you will obtain: k\theta<Y_{(n)}\le\theta. Let Y_1,\cdots, Y_n be a random variable from a given power family distribution. Here, Y_{(n)} is n-th order statistics. Therefore, ... 2 Variance can be evaluated as follows$$ \text{Var}\left[\hat{\theta}_{2}\right]=\text{Var}\left[Y_{(n)}-\frac{n}{n+1}\right]=\text{Var}\left[Y_{(n)}\right]=\text{E}\left[Y_{(n)}^2\right]-\left(\text{E}\left[Y_{(n)}\right]\right)^2. $$First, we calculate \text{E}\left[Y_{(n)}^2\right]. Using the result from here, we obtain$$ ...

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Note that $$\int_{X^{-1}(B)}Y\,\mathrm dP=\int_\Omega Y\mathbf{1}_B(X)\,\mathrm dP={\rm E}[ f(X,Y)]$$ with $f(x,y)=y\mathbf{1}_B(x)$ being Borel measurable from $\mathbb{R}^2$ to $\mathbb{R}$. Since ${\rm E}[f(X,Y)]$ only depends on the distribution of $(X,Y)$ we conclude that ${\rm E}[f(X,Y)]={\rm E}[f(W,Z)]$ or in other words $$... 2 Without loss of generality, we assume that i<j. Then,$$X_j = S_j-S_{j-1}$$entails that$$\mathbb{E}(X_j \mid \mathcal{F}_i) = \mathbb{E}(S_j \mid \mathcal{F}_i)- \mathbb{E}(S_{j-1} \mid \mathcal{F}_i) = S_i - S_i = 0.$$Using the tower property, we see that$$\mathbb{E}(X_i X_j) = \mathbb{E}\big[X_i \mathbb{E}(X_j \mid \mathcal{F}_i) \big] = 0.$$... 1 This is correct, and while it’s pretty cool, it may not be quite as significant as you think. Rare coincidences happen all the time. Imagine I met someone new, and after we exchanged phone numbers, we noticed that they had the same first four digits. Wow! There’s only a 0.01% chance of that. What if our phone numbers had ended in 5463 and 3645, ... 1 There is another way to view the relationship between the gamma distribution and the beta distribution through the Dirichlet distribution. This post (http://math.stackexchange.com/q/190695) talks about exactly how they are related without the Dirichlet distribution, but here is a slightly broader view: Let Z_1, Z_2, \ldots, Z_n be independent random ... 1 No, you cannot make conclusion like that. You are doing it right all the steps except for the regions of u and v. You've obtained the joint PDF of U and V$$ f_{U,V}(u,v)=\frac1{u^2}. $$This is correct. Now for the regions. You have 0\le y\le1, this region is corresponding to 0\le u\le1. It's due to your transformation Y=U. You also have 0\le ... 1 The joint PDF of U and V is$$ f_{U,V}(u,v)=f_{X_1,X_2}(x_1,x_2)\cdot|J|=\frac1{2v}. $$Now for the regions. You have 0\le x_1\le1, this region is corresponding to$$0\le \sqrt{uv}\le1\;\Rightarrow\;0\le uv\le 1\;\Rightarrow\;0\le v\le \frac1u.$$It's due to X_1=\sqrt{UV}. You also have 0\le x_2\le1, this region is corresponding to$$0\le ...

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Hint: Start from the definition and conditional expectations: $$M_X(t):=\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid Y]].$$ We know that $e^{tX}$, conditioned on $Y$, is distributed as a Gaussian with mean $Y$ and variance $1$; so, that inner conditional expectation isn't bad: think about the EGF for such a Gaussian. When you've done that, you ...

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The covariance of $X$ and $Y$ is $E(XY)-E(X)E(Y)$. (The formal definition of covariance is $E((X-E(X))(Y-E(Y)))$, but that is usually, and in this case, harder to work with.) To find $E(XY)$, find the sum $\sum_{(x,y)} xy\Pr(X=x\land Y=y)$. There will be $9$ terms to add up, really only $7$, since $2$ of the terms are $0$, A typical term like the one ...

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If $P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0) \neq 0$, \begin{align*} P(X_n = i_n \mid X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)&=\frac{P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}{P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}\\ &= \frac{P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}{\sum_{i_n}P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}\\ ...

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\begin{align} u & = \sqrt{x}/\theta \\ u^2 & = x/\theta^2 \\ 2u\,du & = dx/\theta^2 \end{align} $$\int_0^\infty e^{-\sqrt{x}/\theta} \, dx = \theta^2\int_0^\infty e^{-u} \Big( 2u\,du \Big) = 2\theta^2.$$ The integral can be done by parts, thus: $$\int u \Big(e^{-u}\,du\Big) = \int u\,dv=uv-\int v\,du = -ue^{-u}-\int -e^{-u}\,du,\text{ etc.} ... 1 The motivation of the method of moments estimate is that it produces a model that has the same first n raw moments as the data (as represented by the empirical distribution). Let Y_1, Y_2, \cdots, Y_n be a random sample, therefore$$ \text{E}\left[Y^n\right]=\frac{1}{n}\sum_{i=1}^n y_i^n.\tag1 $$Let us obtain the first raw moment of the data.$$ ...

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The derivative of the function $u$ defined by $u(x)=\mathrm e^{-\lambda x}$ is such that $u'(x)=-\lambda \mathrm e^{-\lambda x}$. By Taylor formula, for every nonnegative $x$, there exists some $z(x)$ in $[0,x]$ such that $u(0)-u(x)=-xu'(z(x))$, that is, $1-\mathrm e^{-\lambda x}=\lambda x\mathrm e^{-\lambda z(x)}$. When $x\to0$, $z(x)\to0$ hence $\mathrm ... 1 Your answer is correct; the provided answer is the MLE of$\theta + 1$, not$\theta\$. You can see this either through simulation, or by direct calculation: if we let $$s = -\sum_{i=1}^n \log y_i > 0,$$ then $$\hat \theta = -1 + \frac{n}{s},$$ and $$\ell(\hat\theta \mid s) = n \log(\hat \theta + 1) - \hat\theta s = s + n \log \tfrac{n}{s} - n.$$ But ...

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