# Tag Info

15

No, if the individual random variables are continuous and thus their marginal distributions can be described using pdfs, it is not necessarily the case that they enjoy a joint pdf. A standard counterexample to the OP's "claim" is when $X \sim N(0,1)$ and $Z$ is an independent discrete random variable taking on values $\pm 1$ with equal probability. ...

12

No, a very simple counterexample is $(Z, Z)$ where $Z \sim \mathcal{Norm}(0,1)$ where the marginals are standard normal but the joint distribution is concentrated on the diagonal line $y=x$. So the joint distribution do not have a density with respect to the Lebesgue measure on the plane, but it does indeed have a density with respect to the Lebesgue density ...

5

This is laplace distribution, you can either use the CDF given in the wikipedia page or find a proof. If you want to integrate it yourself, the integral will be from $-\infty$ (not $0$) to $x$, and for the negative portion, you'll just substitute $|x|=-x$ and do the integration.

4

Hint: Consider $$(\epsilon_1-\epsilon_3) - (\epsilon_1-\epsilon_2) - (\epsilon_2-\epsilon_3)$$

4

$X$ and $Y^X$ are not independent variables in general. $Y^X$ is a function of the variable $X$ and thus depends on the distribution of $X$. Another way to think about independence is about information, knowing the value of $Y^X$ would give you information about the value of $X$. Re: misconception. The way you have written it, $X$ in $Y^X$ is the same random ...

4

.. each with a specific height that can be measured with infinite accuracy.. Based on this, we could say that height of a single individual, say $X$, is continuous RV, it can be any real number within a plausible range. This makes vector of heights, say $X^N$, a continuous random vector as well. For example, we could have 𝑁=3 people with heights 150 cm, ...

3

The question has nothing to do with measure theory, this is standard basic calculus: If one writes$$T(X_1, X_2,...,X_n) \stackrel{\text{def}}{=} f(X_1,X_2,...,X_n)$$the function $$f\,:\,\mathcal X^n\longmapsto\mathbb R^k$$is mapping $\mathcal X^n$ into a $k$-dimensional space. This means that $f$ and hence $T$ change for each value of $n$, that they are ...

3

Yes, the classical PDF from beginning calculus-based probability is the “Radon-Nikodym derivative” of the probability measure with respect to Lebesgue measure. That more-or-less corresponds to taking the derivative of the CDF to get the PDF. Even though the intro probability class makes it sound like we get CDFs by integrating densities, CDFs come directly ...

2

The next iteration of the rule would be: $$p(x) = \sum_y \sum_z p(x,y,z) = \sum_y \sum_z p(x|y,z) p(y|z) p(z).$$

2

You have $\begin{bmatrix}1 & 0 & -1 \\ 1 & -1 & 0 \\ 0 & 1 & -1\end{bmatrix} \epsilon = \eta$, but this matrix is not invertible, so the vectors of the form $(\epsilon_1-\epsilon_3, \epsilon_1-\epsilon_2, \epsilon_2 - \epsilon_3)$ cannot span all of $\mathbb{R}^3$.

2

If TemperatureGradient and Individual are your fixed effects that you are specifically interested in obtaining inferences for, and you have repeated measures within Trial, then the model should be: HeatingRate ~ TemperatureGradient * Individual + (1|Trial)

2

There are many natural ways this can occur. One is that even a single influential outlier can control the correlations. This situation will be obvious and scarcely needs explaining. To find other such circumstances, work backwards from the absolute differences to construct random variables $(X,Y)$ with the desired properties. Begin with any non-negative ...

2

For the first question, since $Y_{n}$ is of binomial distribution, the probability that $Y_{n}=y$ is $P(Y_{n}=y)={n \choose y}\theta ^y\times(1-\theta)^{n-y}$. The probability that $n-Y_{n}=k$ then is $P(n-Y_{n}=k)=P(Y_{n}=n-k)={n \choose n-k}\theta ^{n-k}\times(1-\theta)^{k}={n \choose k}(1-\theta) ^{k}\times\theta^{n-k}$. Thus, $n-Y_{n}$ is of binomial ...

2

For you second question, since you already know that the mean of $Y_{n}$ is $n\theta$, and $n\sim\mathcal{Poisson}(\lambda)$, the mean of $Y_{n}$ is $E[Y_{n}] = E[n\theta]=\theta E[n]=\lambda\theta$. Or equivalently, $E[Y_{n}]=\sum_{n} n\theta*\frac{\lambda^{n} e^{-\lambda}}{n!}=\lambda\theta$. Similarly, the various of $Y_{n}$ is $\sum_{n} n\theta (1-\theta)... 2 Following from the third line,$E[Y_i-\bar Y|\mathbf X]=(\beta_0+\beta_1X_i)-(\beta+\beta_1\bar X)=\beta_1(X_i-\bar X). When substituted that back, we have \begin{align}E[\hat \beta_1|\mathbf X]&=\sum_{i} \beta_1g_i(\mathbf X)(X_i-\bar X)=\beta_1\sum_i\frac{(X_i-\bar X)}{\sum_j (X_j-\bar X)^2}(X_i-\bar X)\\&=\beta_1\frac{\sum_i (X_i-\bar X)^2}{\... 2 Both notations are correct and you can do them. Once you conditioned on a random variable or vector, you can treat any function of it that doesn't include other RVs as a constant and take it out from the expectation. 1 I love graphical displays. Here are two that illustrate the right hand side of the law of total variance nicely. First, some code for a linear but heteroscedastic regression. set.seed(12345) nsim = 100 X = runif(nsim, 40,120) Y = 1 + 0.3*X + rnorm(nsim, 0, 0.15*X) Cond.Mean = 1 + 0.3*X # Conditional Mean Cond.SD = 0.15*X # Conditional ... 1 A simple example would work best here. Population: Adults aged 20-65 who suffer from heart disease and are patients of a local hospital. Random variable: Heart rate (beats per minute) of a randomly selected adult from this population. The population is a well-defined collection of individuals we are interested in studying. The individuals belonging to this ... 1 You can see it as \hat{\beta}=(X^tX)^{-1}X^ty = (X^tX)^{-1}X^t(X\beta+\varepsilon) =(X^tX)^{-1}X^tX\beta + (X^tX)^{-1}X^t\varepsilon $$So you have that$$ \hat{\beta}=\beta + (X^tX)^{-1}X^t\varepsilon $$And then, it is straightforward to see that$$ \mathbb{E}(\hat\beta) = \mathbb{E}(\beta + (X^tX)^{-1}X^t\varepsilon) = \beta + (X^tX)^{-1}X^t\mathbb{E}(\... 1 This seems to be from an MIT course (or at least I can find the same problem in the problem set 2 of Statistics for Applications Fall 2016). The question reads Problem 2 Statistical models and identifiability For the following experiments, define a statistical model and check whether the pa­rameter of interest is identified. So the question relates to ... 1 If the regression coefficients are estimated using OLS and the variance of the errors is\sigma^2$, then $$\begin{bmatrix} \hat{\beta}_0 \\ \hat{\beta}_1 \end{bmatrix} |X \sim N\Big( \begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} , \sigma^2\begin{bmatrix} n & \sum_i X_i\\ \sum_i X_i & \sum_i X_i^2 \end{bmatrix}^{-1} \Big).$$ where$X=(X_1......

1

Adaptation of this answer. Let $X,Y,Z$ be random variables. Without loss of generality we may assume they have mean zero and variance one, since correlation is invariant under shifting and scaling. Let $\rho_{XY}, \rho_{YZ}, \rho_{XZ}$ be the pairwise correlations (which are equal to covariances since everything has variance one). One can check that $O_{XY} :... 1 This doesn't depend on the possible values but on interpretation and what is measured. If it's counts of something, it's ratio, as if one value is twice as high as another, it really means "twice as much" in a properly interpretable way (actually proper counts are absolute scales, which means that they fulfill the conditions for all lower scale ... 1 For actually generating$n$such values, here are three approaches in R turning uniform pseudo-random variables into your distribution. The first is a simplified inversion of the CDF, putting its two parts together; the second essentially takes an exponential distribution and applies a random sign; and the third is essentially the difference between two ... 1 No, given that$X, Y \sim D$, it is not necessarily the case that$X$and$Y$are independent. For your example, suppose that the joint density of$X$and$Y$has value$2$on the square region$\left[0,\frac 12\right)\times \left[0,\frac 12\right)$and also on the square region$\left[\frac 12,1\right]\times \left[\frac 12,1\right]$. Then,$X, Y \sim U[0,1]$... 1 I would say it is largely a mild curiosity, though there is a real effect Note that the two pairs$(9,3)$and$(1,1)$give both the extreme values of Sample1 and the extremes of the absolute difference; the same cannot be said for the extreme values of Sample2 Even without these two pairs, there is still some relationship similar to the one you observed but ... 1 We can make lots of examples by taking mixtures of compactly supported distributions where the supports are some positive distance apart. For example, let$f_n = 2 \mathbf 1_{[n, n + 1/2]}$so$f_n$is the density of a uniform RV on$[n, n + 1/2]$(I could have done this with translated beta distributions too, among many others). Then if I take$$f =\sum_{... 1 Begin with the following fact:$E[X^m] = \begin{cases} 0, \quad \quad \quad \quad m\ \mathrm{is\ odd} \\ 2^{-m/2}\frac{m!}{(m/2)!}, \quad m\ \mathrm{is\ even} \end{cases}$So,$E[X] = E[X^3] = E[X^5] = 0, E[X^2] = 1,$and$E[X^4] = 3$. For$Y$, use the formula$E[Y] = \sum y P(Y)$Ex:$E[Y] = \sum y P(Y) = (-\sqrt{3})(1/6) + (\sqrt{3})(1/6) + (0)(⅔) = 0$... 1 Your question shows some confusion! A sample is usually represented by a sequence of random variables iid$X_1, X_2, \dotsc, X_n, \dotsc$, where each random variable is a function from some sample space$\omega \mapsto X(\omega)$, the argument$\omega$often omitted from the notation. So a realization is only one value, always. So$n$in the CLT refers to ... 1 I think the most likely origin is that if you take any moment formula from$\mathrm{Bin}(n,q)$and you replace the parameters$n$and$q$in that formula with negative values$-\alpha$and$-\theta$, then the result will be the equivalent moment formula for a Negative Binomial distribution with parameters$\alpha$and$\theta\$. For short, it looks like as if ...

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