I am doing some reading and am seeing phrases such as the following.  For context, I am learning about Bayes Inference and prior/posterior distributions so theta is the parameter:     In such problems, we shall refer to the distribution indexed by θ for   the other random variables of interest as the conditional distribution   for those random variables given θ.   ...     In many problems, such as Example 7.2.1, the observable data X1, X2, .   . . are modeled as a random sample from a univariate distribution   indexed by θ.   ...     In these cases, let f (x|θ) denote the p.f. or p.d.f. of a single   random variable under the distribution indexed by θ.   What does "indexed" mean and how do I conceptualize it?
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Does a family of a distribution have a different definition for statistics than in other disciplines?  In general, a  is a set of curves, each of which is given by a function or parametrization in which one or more of the parameters is varied. Such families are used, for example, to .  For statistics, a family according to  is the result of varying the shape parameter. How then can we understand that the gamma distribution has a shape and scale parameter and only the generalized gamma distribution has, in addition, a location parameter? Does that make the family the result of varying the location parameter? According to @whuber the meaning of a family is implicitly A "parameterization" of a family is a continuous map from a subset of ℝ$^n$, with its usual topology, into the space of distributions, whose image is that family.  What, in simple language, is a family for statistical distributions?  A question about relations among of the statistical properties of distributions from the same family has already generated considerable controversy for a  so it seems worthwhile to explore the meaning.  That this is not necessarily a simple question is born out by its use in the phrase , which has nothing to do with a family of curves, but is related to changing the form of the PDF of a distribution by reparameterization not only of parameters, but also substitution of functions of independent random variables.
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I am working on a problem with the following set:  $$S =  \{\varnothing,\{\varnothing\}\}$$  My solution was: $P(S) = \{\varnothing, \{\varnothing\},\{ \varnothing , \{ \varnothing \}\}\}$, but the solution on the textbook shows: $P(S) = \{ \varnothing , \{ \varnothing \}, \{\{\varnothing\}\}, \{ \varnothing , \{ \varnothing \}\}$  Where does the $\{\{\varnothing\}\}$ come from?  Thanks
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Suppose that in a Bayesian framework we have observed data $D$, using independent prior distributions on the parameters of the model, denoted by $\theta_1, \theta_2$. Then, the joint posterior distribution of $\theta_1, \theta_2$ can be written as: $$ P(\theta_1, \theta_2 \mid D) \propto P(D \mid \theta_1, \theta_2)p(\theta_1)p(\theta_2) $$  One way to obtain estimates from the posterior is use . This sampler requires simulating from the full conditional distributions for both $\theta_1$, $\theta_2$. One relation for the full conditional that caught my eye in a book is the following: $$ P(\theta_1\mid D, \theta_2) \propto P(D \mid \theta_1, \theta_2)p(\theta_1) $$  I am wondering how exactly this proportional relation is derived. Would anyone have any ideas? It seems that there are extra assumptions here that are implicit.
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$Y_{1...n}\sim \operatorname{Bin}(1,p)$, iid,  and I need to find an unbiased estimator for $\theta=\operatorname{var}(y_i)$.  I did some calculations and I think that the answer is $p(1-p)-\frac{p(1-p)}{n}$   Is this correct? If not, how can I find an unbiased estimator?
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My notes are rather light when it comes to this topic. I understand that the bayesian estimator, defined as (for sample space $\hat{x}$):  $E[\Theta | \hat{x}] = \int_{ \forall \Theta}yf_{\theta|\hat{x}}(y|\hat{x})dy $ (ie. the mean of the posterior distribution).  You can then look at loss functions where the form of the loss function determines that of $\theta^{'}$ (the estimator of $\theta$). By setting the loss function as a quadratic, absolute error or zero-one you get $\theta^{'}$ as the mean of the posterior, the median of the posterior and the mode of the posterior respectively.   The first two proofs make sense to me but it is the 3rd that I am not sure about. This is my understanding:  $E[l] = \int l(\theta^{'}, \theta)f(\theta|\hat{x})d\theta = \int (\theta^{'} - \theta)f(\theta|\hat{x})d\theta $  If $\theta^{'} = \theta \Rightarrow E[l] = 0$  If $\theta^{'} \neq \theta \Rightarrow \int (\theta^{'} - \theta)f(\theta|\hat{x})d\theta = 1 $ (as integrating over whole domain of $\theta$).  This proof obviously isn't rigorous, (and it could be incorrect too). What I can't see regardless is how this can be regarded as the 'mode' of the posterior distribution.   Thanks in advance!
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Given a random variable $X$ which follows a normal distribution $X \sim (\mu,\sigma^2)$ with unknown mean and known variance. Say we have $n$ observations $y_i$ where $1\le i \le n$ and each $y_i$ is sampled from a normal distribution $Y_i \sim N(x,\sigma_i^2)$ (not i.i.d.) where $x$ is sampled from $X$ and it is the same for all $\{Y_i\}$, but $\sigma_i$ is different for different $Y_i$.  The question is: how to compute the posterior distribution $X$ given the observations p(X|{y_i})? The question is different from the answer listed in  because $y_i$ is not sampled from i.i.d. distribution.
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Suppose that we observe the discrete random variable $ X = (X_1, \dotsc , X_n)$ with state space S, whose distribution we do not know but we are assuming that its joint p.m.f. belongs to a known family $\{f_θ : θ ∈ Θ\}$. We derive a sufficient statistic $T(X)$.  Is it correct to write $\Bbb E[X]$? If not, why? Would it be more correct to write $\Bbb E_{\theta}[X]$?
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So I'm reading notes on Gaussian Processses, and came across  syntax $p(\mathbf{y} \mid \text{stuff}) = N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ for multivariate normal distribution, and I'm not exactly sure how to decipher it the $\mathbf{y} | $ bit. I suppose it doesn't mean "mean $\mathbf{y}$ is conditional on zero vector", because I don't think that makes sense. Assuming this is standard notation for something and I'm just unfamiliar with it, what it means? E.g. how it differs from $N(\mathbf{0}, \mathbf{K})$?
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I am trying to do an excerise on likelihood functions and score vectors and I am having some trouble extending it to parameter vectors. The question is stated below. Suppose we have two independent samples: $X_{1}, ..., X_{n}$ are iid Exponential random variables with mean $θ_{1}$ and $Y_{1}, . . . , Y_{n}$ are iid Exponential random variables with mean $θ_{2}$. Find the likelihood and log likelihood functions for the parameter vector $θ = (θ_{1}, θ_{2})^{T}. $ In this case would the likelihood function be $$\prod{θ e^{-θ x_{i}} = θ ^ne^{-θ\sum x_{i}}}$$ (same as one with one parameter)? Or would it be the product of the two functions as shown below: $$\prod{θ_{1} e^{-θ_{1} x_{i}} θ_{2}e^{-θ_{2}y_{i}} = θ_{1} ^ne^{-θ_{1}\sum x_{i}}} θ_{2}^{n}e^{-θ_{2}\sum y_{i}}$$ Or would it be something different? Thanks!
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Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that  $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?  If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?),  I was  the  that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.  So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.  p.17  seems to suggest otherwise. idk  Help please?  Also this:    or    So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
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I have 2 i.i.d Uniform random variables $\theta_{1}$ and $\theta_{2}$ on $[0, 2\pi)$. Here the problem pushes us to consider that all operations are made modulo $2\pi$. My question is how to obtain the joint PDF of $\theta_{1} - \theta_{2}$ and $\theta_{1}$ and then obtain the conditional PDF of $\theta_{1}$ given $\theta_{1} - \theta_{2}$??. For the second part It can be deduced from the first by writing it as the joint PDF divided by the PDF of $\theta_{1}$. But for the first part, A common idea that I saw is to derive the $\text{CDF}(x, y) = P(\theta_{1} - \theta_{2} \le x$, $\theta_{1} \le y$) and then perform derivations respect to $x$ and $y$. Is this a right approach? If yes, how to proceed with CDF especially with our uniform distribution restrictions?
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Some sources say likelihood function is not conditional probability, some say it is. This is very confusing to me.  According to most sources I have seen, the likelihood of a distribution with parameter $\theta$, should be a product of probability mass functions given $n$ samples of $x_i$:  $$L(\theta) = L(x_1,x_2,...,x_n;\theta) = \prod_{i=1}^n p(x_i;\theta)$$  For example in Logistic Regression, we use an optimization algorithm to maximize the likelihood function (Maximum Likelihood Estimation) to obtain the optimal parameters and therefore the final LR model. Given the $n$ training samples, which we assume to be independent from each other, we want to maximize the product of probabilities (or the joint probability mass functions). This seems quite obvious to me.  According to , "likelihood is not a probability and it is not a conditional probability". It also mentioned, "likelihood is a conditional probability only in Bayesian understanding of likelihood, i.e., if you assume that $\theta$ is a random variable."  I read about the different perspectives of treating a learning problem between frequentist and Bayesian.   According to a source, for Bayesian inference, we have a priori $P(\theta)$, likelihood $P(X|\theta)$, and we want to obtain the posterior $P(\theta|X)$, using Bayesian theorem:  $$P(\theta|X)=\dfrac{P(X|\theta) \times P(\theta)}{P(X)}$$   I'm not familiar with Bayesian Inference. How come $P(X|\theta)$ which is the distribution of the observed data conditional on its parameters, is also termed the likelihood? In , it says sometimes it is written $L(\theta|X)=p(X|\theta)$. What does this mean?  is there a difference between Frequentist and Bayesian's definitions on likelihood??  Thanks.    EDIT:  There are different ways of interpreting Bayes' theorem - Bayesian interpretation and Frequentist interpretation (See: ).
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How is the notation $X\sim N(\mu,\sigma^2)$ read? Is it $X$ follows a normal distribution? Or $X$ is a normal distribution? Or perhaps $X$ is approximately normal..  What if there are several variables that follow (or whatever the words is) the same distribution? How is it written?
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I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) &gt; Var(X)$ where $i &gt; 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.  Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?
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I am having an issue finding a best unbiased estimator for $\theta$. Any help is appreciated. Let $X_1, ..., X_n$ be a random sample from a population with pdf: $f(x\mid\theta)=\frac{1}{2\theta}$ $-\theta&lt;x&lt;\theta,\, \theta&gt;0$. I understand that $T(x) := |X_n|$ is the sufficient statistic since by the factorization theorem we have: $$f(x\mid \theta) = \left(\frac{1}{2\theta}\right)^n \prod_{i=1}^n I[|x_i|&lt;\theta]$$ I think my main issue is showing that this sufficient statistic is also a complete sufficient statistic. Can somebody please aid me in this?
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In my statistics class, we're just beginning to talk about (point) estimation. I have a small question that might actually be due more to notation/definition than anything very conceptual:  Say you have a population $P$ of students in a classroom and you want to find the mean height by randomly selecting $n$ students $s_1,...,s_n$ in $S\subseteq P$ to measure. My book calls $X_1,...,X_n$ a random sample. I'm just a little confused as to what $X_i$ is defined to be (what $X_i$, as a random variable, is mapping from). In this example, my book would say {$X_1,...,X_n$} are the respective heights of the $n$ randomly selected students. Just to make sure I have the right picture in my head: $S$ is the ("first") sample space and $X:S\to\mathbb{R}$ is a random variable ($X$ is the "height" function defined on $S$). We denote $X_i$ as the restriction of $X$ to the $s_i$-th student in $S$. That is, we define $X_i:${$s_i$}$\to\mathbb{R}$ so $X_i(s_j)$ is not defined for $i\neq j$, and:   $X_1(s_1)=X(s_1)=x_1$  $X_2(s_2)=X(s_2)=x_2$  $.$  $.$  $.$  $X_n(s_n)=X(s_n)=x_n,$ where $x_1,...,x_n$ are the respective heights of the $n$ students. Then $Range(X)=${$X_1(s_1),...,X_n(s_n)$}$=${$X_1,....,X_n$} is considered the new sample space that we define the pmf/pdf $f$ on, where $f:\mathbb{R}\to\mathbb{R}$ and $f\circ X(s_1)=f(x_i)$.   Is this the right idea?
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I am preparing a presentation on statistics. I want to include a slide on Bayes' Theorem which says:  p(\theta | D) = \frac{p(D|\theta) p(\theta)} {p(D)}   I want to explain each term using arrows. How can I do this?  I tried to follow this  but my main problem is that I have a fraction. How can I split the fraction?   is exactly what I want to do. Without the last line.