Tag Info

Hot answers tagged

2

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 $$ ...


2

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 ...


1

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 ...


1

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 ...


1

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)}\\ ...


1

\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. $$ ...


1

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible