# Tag Info

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Independent and identically distributed. If you ask for IID $X$ and $Y$, as others have noted this is not possible. See the answers to this question. If you are happy to drop either independence between $X$ and $Y$ or them having the same distribution, then there's hope. Same distribution. If you allow for dependent but identically distributed variables, ...

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Consider the following example: $$X\sim\text{Unif}(0, 1) \\ Y = 1-X$$ $X$ and $Y$ are identically distributed as the standard uniform distribution, and $X-Y = 2X-1$, so $X-Y\sim\text{Unif}(-1, 1)$. Note that this example relied on $X$ and $Y$ being dependent, identically distributed random variables. It is impossible for two independent, identically ...

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Every moment/standardised moment is expressed in terms of some kind of expected value. For example, you've expressed the variance using expected value. But you couldn't do it using only the expected value of $X$, i.e. $E[X]$, since you cannot express $E[X^2]$ using $E[X]$ only. Similarly, the skewness can be expressed in terms of expectations as follows: $$E\... 4 This is the set C_{m,n} of matrices that have at most one nonzero entry in each column. Let us rephrase the issue in the question as a property of an m\times n matrix A: \mathcal{P}(A): for all independent random variables x=(x_1,\ldots,x_n) defined on a given space \Omega, z=Ax is independent. My assertion requires two demonstrations: ... 3 For orthogonal columns, in general it's not. Let$$A=\begin{bmatrix}1 &-1\\1&1\end{bmatrix}$$This results in z_1=x_1-x_2,z_2=x_1+x_2. These two aren't independent in general. For example, assume x_i are Bernoulli RVs, then if z_2=2, z_1 is definitely 0, which means there is dependency. For the linear independence, see @David's counter-... 0 So first: When you define a random variable X_{n+1} to depend on the value of a different random variable X_n you are effectively defining the conditional probability P(X_{n+1}|X_n) as long as P(X_n) was well defined this defines a probability distribution P(X_{n+1}, X_n) over the two variables with the corresponding outer product \Omega and \... 4 Your first question is key, so let's focus on it. You are concerned about a bivariate random variable (X_{n-1},X_n) with a probability distribution somehow defined by giving X_{n-1} a distribution and then defining the distribution function of X_n in terms of the random variable X_{n-1}. There are many subtleties involved, so I write the following ... 1 Definitions Definition 1: If \mathcal F \subseteq \mathcal G are two \sigma-fields, and X a \mathcal G-measurable integrable random variable, then \mathbb E[X | \mathcal F] is defined as any \mathcal F-measurable random variable Y, such that \mathbb E[Y;A]=\mathbb E[X;A] for every A \in \mathcal F. Here \mathbb E[X;A] is a notation for ... 2 It's not analytically available. You can use numerical approaches/software. What you want to find is actually inverse CDF of quantile function in other words. In Matlab you can use gaminv, in R, you can use qgamma, or in python you can use ppf in scipy.stats. It's not easy to find a table but here is one with unit scale, with varying shape parameters (upto ... 0 Any two sample space can be combined by the operation of direct sum. That's generally denoted by the symbol \oplus, although the simple addition symbol is sometimes used when it's considered obvious that the direct sum is intended. However, in this case, it appears that regular addition is meant. The coin flips are being represented by the integers {0, 1}... 8 It's a good question, interpreted in the following way: the random variable X determined by the coin experiment has a sample space of \Omega_1=\{\text{Heads},\ \text{Tails}\} while the random variable Y determined by the roll of a die has a sample space \Omega_2 consisting of the six possible stable orientations of the die on the table. These ... 0 Remember that X and Y will be the outcome of the roll of a dice/ flip of a coin. So if I understand your question correctly, you will ad the value you get from your first die roll (say 2) to the value of the coin toss (say 0). Resulting in a Z of 2 for example. 4 Surely they can have different sample spaces (or supports) but you can add them and generate another RV. If you assume independence between the two, the resulting mass function can be calculated by simply convolving the two. But, a more primitive solution is to list all possible scenarios:$$\begin{align}&P(Z=1)=P(Y=0)P(X=1)=1/12\\&P(Z=2)=P(Y=0)P(X=...

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Here's the way I would approach it: If $A$ is the number of marbles chosen by the first process, $B$ is the number of marbles chosen by the second process, and $k$ is the number of marbles that were chose by both processes, then the total number of marbles chosen is $A + B - k$. I'm assuming we know $A$ and $B$, and what we're trying to estimate is $A + B ... 0 For$M_n^{(1)}$we must first check that the process is integrable. Note that$\theta\mapsto |e^\theta+e^{-\theta}|$is an even function which is increasing on$(-\infty,0)$and decreasing on$(0,\infty)$and hence attains its maximum$2^{-n}$at$\theta=0. Now, \begin{align} \mathbb E[|M_n^{(1)}|] &= \mathbb E\left[\left|\frac{2e^{\theta S_n}}{(e^\... 2 The\theta's$are the same, but the$c_i's$are not. The fact that each$X_i$has its own$c_i$is what makes the distributions not (necessarily) identical. 1 You can either condition on$X$or condition on$Y$. But you can't condition on$f_Z(Z)$as that is the quantity that you would like to find. If you would like to condition on$X$assuming that$X$and$Yare independent, \begin{align} Pr(Z \le z) &= Pr\left(\frac{X}{Y} \le z\right) \\ &=\int_0^\infty Pr(Y \ge \frac{X}{z}|X=x)f_X(x) \, dx \\ &=... 2 Clearly, takeX \sim \mathcal{MVN}(0, \Sigma)$where$\Sigma$is an equicorrelation matrix with 1 on the diagonal and$1>\rho>0$outside the diagonal. Then$X$is infinitely exchangeable, but the components are not independent. With negative$\rhoX$is weakly exchangeable, see When are observations not weakly exchangeable?. 1 Consider a function$f: \mathcal X \rightarrow \mathcal X$. How would you distinguish between$f$and a particular realization of it,$f(x)$? In general this is completely clear, the function is an object that lives on some abstract space while$f(x)$is some value in$\mathcal X$. A random variable is the exact same -$Y$is in reality a function from a ... 5 The support is the rectangle$\mathcal{R}=[0,3]\times[0,2]$, and for$P(X>Y)$, you'll integrate the area under$Y=X$line: $$\int_{\mathcal R} f_{X,Y}(x,y)dydx=\int_{0}^1\int_0^x \frac{1}{9} dy dx+\int_1^2\int_0^x\frac{2}{9} dydx$$ 0 Here is a classical counterexample: Let$\Pr(z_n=0)=(n-1)/n$and$\Pr(z_n=n)=1/n$. Then, for any$\epsilon$, $$\Pr[|z_{n}-0|<\epsilon]=\Pr(z_n=0)=(n-1)/n\rightarrow 1,$$ i.e.$z_{n}\to_p0$. However, $$E(z_n-0)^2=n^2\cdot 1/n\rightarrow\infty,$$ so that$z_{n}\nrightarrow_{\mathrm{m.s.}}0$. The reason you cannot just apply the expected value operator ... 1 Stochastic processes in continuous time are complicated, because they in a sense simultaneously live in two spaces separately. More abstractly, a stochastic process is just a random variable$X$taking values in a function space, for example the space of continuous functions with input$[0,T]$taking values in$\mathbb R$, we typically denote this space as ... 0 Based on your graph, it appears that the association between time and the number of eggs varies across houses. To test this model, your syntax would be: Model2 <- glmer.nb(Eggs ~ temp + Pp + NDVI + Date + (Date|house), data=HueRC) The (Date|House) syntax allows the association between Date and Eggs to vary across Houses. I would suggest that you test ... 3 I think you used the Cholesky decomposition where$L$is a lower triangular matrix, meaning$b=0$. I hope this helps. 1 The converse is also true and you can write the joint pmf as the multiplication of the two marginal pmfs. 3 It means that you will need to confirm that this indeed is a global maximum by showing that the second derivative of the likelihood function and that point is negative and to test the values of the likelihood at the boundaries$θ→0^+$and$ θ→∞$as whuber pointed out in his comment. 0 Theoretically speaking the outcomes of an experiment (experiment = a random procedure) can be numerical (i.e rolling a die) or can be mapped to numbers by the designer (i.e flipping a coin, with outcomes 1=head and 0=tail). This numerical representation of the outcomes defines the random variables. In these examples, we can tell that there is some chance ... 1 Welcome to this community! The following answer comes from the point of view of probability as a generalization of logic, and as an expression of a person's degrees of belief. First of all I'd like to emphasize that independence is not a physical property of objects or quantities. It is a property of a person's degree of belief about those quantities. It ... 0 Welcome to the site, roc. It may be possible to employ a mixed modeling framework to your data. How many different species did you record? If there are 10-15 or more, then you could have the grouping variable be species. This would indicate that you believe that vocalization frequency is more correlated within species than it is across different species. ... 2 The likelihood in the linked question is $$L(\theta)=P(X=x|\theta)=\theta^x(1-\theta)^{1-x}$$ Here$X$is a Bernoulli RV, and the likelihood function is actually a probability mass function in terms of$x$. So, you can sum wrt$x$and obtain$1$. However, likelihoods are typically for observing/estimating the parameter when the data is given, i.e.$x$'s. So, ... 1 No, you cant, at least not without further restrictions. One restrictions that could do is that the range of$X$is a finite interval. One simple example, see the coments for another example. Let$X$be a discrete random variable such that $$\DeclareMathOperator{\P}{\mathbb{P}} \P(X=\pm a)=p/2, \qquad \P(X=0)=1-p.$$ Then$\DeclareMathOperator{\E}{\...

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