8

I'm assuming you are referecning to Inverse Transform Sampling method. Its very straight forward. Refer Wiki article and this site. Pareto CDF is given by: $$ F(X) = 1 -(\frac{k}{x})^\gamma; x\ge k>0 \ and \ \gamma>0 $$ All you do is equate to uniform distribution and solve for $x$ $$ F(X) = U \\ U \sim Uniform(0,1) \\ 1 -(\frac{k}{x})^\gamma = u \\ x =...


6

Since$$p(\theta_1,\ldots,\theta_{k-1})\propto (1-\theta_1-\cdots-\theta_{k-1})^{\alpha_{k}-1}\prod_{i=1}^{k-1} \theta_i^{\alpha_i-1}$$over the $\mathbb R^k$-simplex, $$\mathfrak S = \left\{(\theta_1,\ldots,\theta_{k-1})\in\mathbb R^{k-1}_+\,;\,\sum_{i=1}^{k-1} \theta_i\le 1\right\}$$ integrating out $\theta_{k-1}$ produces the marginal density of $(\theta_1,\...


5

The same information can be represented by both column and row vector. For example, consider the following vector: $$\mathbf{x} = [ x_1, x_2, x_3]$$ this vector can be transposed as: $$\mathbf{x}^T = \begin{bmatrix}x_1 \\x_2 \\ x_3\end{bmatrix}$$ Both of the above vectors contain the same information. One is just a transpose of the another. $$[ x_1, x_2, x_3]...


5

Let's not try to generalize about "questions like these," but to answer the specific question at hand, using the definition of an F-distribution in terms of chi-squared distributions, as mentioned in a Comment by @whuber. [While moment generating functions and bivariate transformations with Jacobians are often useful, they are not needed in this ...


5

One indication of the difference between (a) $x=5$ successes in $n=10$ tosses and (b) $x = 50$ successes in $n = 100$ tosses lies in the different widths of the 95% confidence intervals (CIs) in the two cases: Wald CIs. Even though the Wald Ci is not the most accurate kind of CI, it is perhaps the most familiar, so I will start by making 95% Wald CIs for ...


5

Since $S_N$ is the sum of $N$ many i.i.d. $\exp(\lambda)$ random variables, therefore $S_N \sim \Gamma(N , \lambda)$ . Now, $N \sim \text{Bin}(n , p)$. Therefore, $S_N~|~N = k~ \sim \Gamma(k , \lambda)$ . Thus, $S_N$ is a mixture of a Gamma r.v., and a Binomial r.v., and it's PDF will be : \begin{align*} f_{S_N} (x) &= \sum_{k = 0}^n f_{\Gamma}(k , \...


4

If I translate in mathematical terms what I believe you said, we have that if we designate by by $(a_{i,j})_{1\le i,j\le n}$ the cases of your grid and $X$ a person, then you have the data of the probability that person $X$ pop up in case $a_{i,j}$. i.e., you have access to the $n^2$ numbers $(p_{i,j})_{1\le i,j\le n}$ such that $$\mathbb{P}(X \text{ is in ...


4

The Conway-Maxwell-Poisson model has recently been shown to handle arbitrarily small underdispersion: https://arxiv.org/abs/2011.07503. For example, it is possible to have a mean of 15 and a variance of 2, say, by selecting the dispersion parameter large enough. In the extreme limits, it is even possible to have a mean of 15 and 0 variance, or a mean of 15.2 ...


4

There seems to be a smooth dependence of variance on observation index, so you could try a joint modeling approach, see for instance Articles that work with covariates for mean, variance, and correlation simultaneously. Maybe also look into if there is autocorrelation (show us a plot!), and tell us what your data represents, and how it was obtained. There ...


4

Graphs in R per @whuber's Comment: set.seed(1117) par(mfrow=c(1,3)) w = rnorm(10^6, 150, 15) summary(w); sd(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 77.08 139.86 150.01 150.02 160.15 221.71 [1] 1.000814 hdr1 = "W ~ NORM(150, 15)" hist(w, prob=T, br=30, col="skyblue2", main=hdr1) curve(dnorm(x, 150, 15), add=T, col=&...


4

Here's a sketch of a solution. First, ignore the scale factor of $\sigma = 2^{-n+1}$ because it can be restored at the end. Because the characteristic function of a Gamma$(1/2)$ variable like $X$ or $Y$ is $$\psi_X(t) = \psi_Y(t) = (1 - it)^{-1/2},$$ the characteristic function of $X-Y$ (a version of the Fourier Transform of its density) is $$\psi_{X-Y}(t) =...


3

By definition, $$F(x_1,\ldots,x_n) = \Pr(X_1\le x_1,\ldots,X_n\le x_n).$$ Assuming these variables are absolutely continuous means there is a density function $f$ for which $$F(x_1,\ldots, x_n) = \int^{x_n}\cdots\int^{x_1} f(t_1,\ldots, t_n)\,\mathrm{d}t_1\cdots\mathrm{d}t_n.$$ Observe two things: The Fundamental Theorem of Calculus, applied $n$ times to ...


3

Why do you expect there to be a connection? Say (without loss of generality) the only thing you know about the distribution of a random variable $X$ is that its median is zero. To simplify lets assume a continuois distribution with cdf (cumulative distribution function $$\DeclareMathOperator{\P}{\mathbb{P}} F(x)= \P(X \le x)=\frac12 $$ This restrains the ...


3

Physical experiments with mechanized tossing of coins have shown that it is very difficult to make an actual coin that is heavily biased. But it is easy to make unfair dice (by shaving edges or embedding lead weights), so I'll consider a die with 3 faces labeled H and three labeled T. Suppose I've had a chance to look at a die and to know how it will be ...


2

Loosely speaking, the Glivenko-Cantelli theorem says that the empirical distribution converges to the population distribution. This is how the world should work: if we keep drawing from a population, we should be pretty close to the population. Maybe it takes ten draws to get "close" and maybe it takes ten-trillion-bazillion draws, but we should ...


2

Let's offer a "dissenting" view: Ratios and inverses of random variables can be fine in the following sense: It may be the case that in many cases they do not possess moments But it is also the case that in many cases they result in recognizable, "named" and exhaustively studied distributions. ...and there is distribution-life beyond ...


2

Using Bayes' rule you can define $$f(\mu|x,\sigma^2) = \frac{f(\mu,x|\sigma^2)}{f(x|\sigma^2)} = \frac{f(x|\mu,\sigma^2)f(\mu|\sigma^2)}{f(x|\sigma^2)}$$ Now $f(x|\mu,\sigma^2) = N(\mu,\sigma^2)$ $f(\mu|\sigma^2) = f(\mu) = N(m,s^2)$ $f(x|\sigma^2)$ is a constant that you don't need Therefore you can see that $$f(\mu|x,\sigma^2) \text{ is proportional to ...


2

These are indeed the same distribution. I've found several parameterizations in use: The Nelson (1991) parameterization is $$f(z) = \frac{\nu\cdot\exp\left[ -\frac{1}{2}|z/\lambda|^{\nu} \right]}{\lambda 2^{(1+1/\nu)}\Gamma(1/\nu)}.$$ To get the parameterization used on Wikipedia, let $\nu=\beta$ and $\lambda=\alpha 2^{-1/\beta}$, and add a mean parameter $\...


2

Your goal is ultimately to roll a k-sided die given only a p-sided die, without wasting randomness. In this sense, by Lemma 3 in "Simulating a dice with a dice" by B. Kloeckner, this waste is inevitable unless "every prime number dividing k also divides p". In your case, p is 2 (since you're gathering binary or base-2 data) and k is 72 (...


2

Using what @TMat said, I got this code to work using numpy img = Image.open("path/to/img") img = np.array(img)[:,:,0] prob = (np.array(img)/np.sum(img)).flatten() x,y = np.meshgrid(np.arange(0,img.shape[0]),np.arange(0,img.shape[1])) x,y = x.flatten(),y.flatten() xy = np.vstack((x,y)).T xy_indices = np.arange(len(xy)) choices = np.random.choice(...


2

One obvious way to construct discrete distributions on the set $\mathbb{Q}_* \equiv \mathbb{Q} \cap [0,1]$ is to do so via sequences of values on that set that cover that set. Suppose you have some arbitrary surjective function $H: \mathbb{N} \rightarrow \mathbb{Q}_*$, which can be interpreted as a sequence on $\mathbb{Q}_*$ that fully covers $\mathbb{Q}_*$ ...


2

I will share three general techniques for working with density functions that are remarkably simple and useful: normalization, scaling, and identifying symmetries. (A fourth general technique, recentering, often is useful but isn't applicable in this situation.) By following these principles, along with a healthy application of the "principle of ...


2

I will give you some clues toward a proof and, I hope, some idea why this equality is of interest. Definition and simulation. First, the distribution $\mathsf{F}(m,n)$ is defined as the distribution of the ratio of two independent chi-squared random variables, each divided by its degrees of freedom (so that numerator and denominator both have mean $1).$ It ...


2

Let's denote by $T$ the number of positive tests in your sample, and by $T_1, T_2$ and $T_3$ the number of positive tests in categories 1, 2 and 3. Then $T = T_1 + T_2 + T_3$. First, let's consider that $x_1, x_2, x_3$ are fixed. Then $T_1, T_2$ and $T_3$ are independent binomial random variables : $$\begin{array}{ccc} T_1 & \sim & \mathcal{B}(x_1, ...


2

On a fundamental level, Monte Carlo is an approach to approximating integrals. It is not intrinsically a method for forecasting. It becomes a method for forecasting when your model defines a certain integral which corresponds to the forecast that you are trying to approximate. Because you have not explicitly defined a model, there is no way to tell if your ...


2

$\int_a^b f(t) X \, dt = \left(\int_a^b f(t) \, dt\right)\cdot X$. (This holds for each realization of the random variable $X$.) The integral $\int_a^b f(t) \, dt$ is just a number, say $c$, and the distribution of $cX$ is $N(0, c^2)$.


2

Judging unfairness. A common way to judge whether faces on a die are equally likely is to use a chi-squared goodness-of-fit test. Suppose you roll a fair die 600 times and count the numbers of 1s, 2s, ..., 6s observed. We can simulate this by using R: set.seed(116) # for reproducibility d = sample(1:6, 600, rep=T) x = tabulate(d); x [1] 103 116 88 98 ...


2

Are you looking for something like this? Write $$X_{N,i} \equiv X_{i} - \frac{1}{N},\quad X_{N+1,i} \equiv X_{i} - \frac{1}{N+1}$$ Let $$Z_N = \frac{1}{2}\sum_{i = 1}^{N}\bigg|X_{i} - \frac{1}{N}\bigg| = \frac{1}{2}\sum_{i = 1}^{N}\bigg|X_{N,i}\bigg|$$ $$Z_{N+1} = \frac{1}{2}\sum_{i = 1}^{N+1}\bigg|X_{i} - \frac{1}{N+1}\bigg| = \frac{1}{2}\sum_{i = 1}^{N+1}\...


2

$Z$ is the mean of $\frac{1}{2}|Y_i -1|$ with $Y_i \sim Exp(1)$ Note that $|X_i - \frac{1}{N}| = \frac{1}{N} |N X_i - 1|$, and $NX_i \sim Exp(1)$. So you can consider the following equivalent problem: Let $Y_{1}, Y_{2}, \ldots Y_{N}$ be non-negative continuous i.i.d random variables such that the probability density function of each $Y_{i}$ is given as \...


2

Firstly, let's start by fixing your estimator. Although $p(1-p)=p^2$ in the special case where $p=\tfrac{1}{2}$, in the context of parameter estimation, we don't know the parameter $p$, and so we don't know that $p(1-p)=p^2$. Consequently, the plug-in variance estimator is $\tilde{\tau}^2_{n} = \tilde{p}_n (1-\tilde{p}_n)$, not $\tilde{p}_n^2$. I will ...


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