# Tag Info

4

Let's write the calculus formally. $$f(x) = \begin{cases} 160x^{-6} & x\ge2\\ 0 & x< 2 \end{cases}$$ To get from $f(x)$ to $F(x)$, integrate: $$F(x) = \int_{-\infty}^{x}f(t)dt.$$ But we can break up the integral into an integral from $-\infty$ to $2$ and another from $2$ to $x$. $$F(x) = \int \limits_{-\infty}^{x} f(t) \ ... 4 Letting Y_i \equiv - 2 \ln R_i and noting that R_i \sim \text{Rayleigh}(1) you have the CDF:$$ \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y_i \leqslant y) &= \mathbb{P}(-Y_i/2 \geqslant -y/2) \\[6pt] &= \mathbb{P}(e^{-Y_i/2} \geqslant e^{-y/2}) \\[6pt] &= \mathbb{P}(R_i \geqslant e^{- y/2}) \\[6pt] &= \exp \Big( -\frac{e^{...

3

$N(x|\mu, \sigma)$ combines the two notations: $x \sim N(\mu, \sigma)$ and $p(x| \mu, \sigma)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma$.

3

I think this was mistakenly migrated to CV, and so I'm going to answer it as I would have on SO, with R code. R has an approximation functional that returns a function object, approxfun. The result can be given an x value in the range of the original x argument and will return a y value. myfun <- approxfun(x,y) png(); plot(-20:10, myfun(-20:10) ) ; ...

2

In literature, normalization means integrating to $1$, not having a max value equal to $1$. So, joint or univariate densities are already normalized. For the nomenclature, for the function you have, I think max-normalized joint density would be a better name for it. However, what you do is just scaling your joint PDF so that it hits $1$ at its maximum. ...

2

The normal curve, otherwise known as the bell curve or Gaussian, is a mathematical model. If you want the specifics it is below, but this equation is not necessary to understand the rest, just skip to the point by point reply. $$f(x) = ae^{-(x-b)^2 / 2c^2}$$ a is a scalar multiplier, b is the central position and c is the spread or standard deviation. It ...

2

You omit $n$ in the expression. Your PDF can be written as $$f_X(x)=\frac{1}{\sqrt{2\pi}(3/n)}\exp(-(x-2)^2/(2(3/n)^2))$$ where $\sigma_2=3/n$.

2

The professor is not using a pdf, the professor is using a likelihood. Consider the following pdf: $$f(x_i|\mu)=\frac{1}{\pi}\frac{1}{1+(x_i-\mu)^2},\forall{x_i}\in\Re.$$ You are correct that $\Pr(x_1=5)=0$ as there is no area. However, the professor is not using a density, the professor is using a likelihood. Consider the following likelihood for the ...

2

For an affine transformation of a random vectors, the following rules apply. Let $\mathbf{X}$ be the $n\times 1$ column vector with random variables $X_1, X_2, \ldots, X_n$ as its elements. Let $\mathbf{a}$ and $\mathbf{b}^T$ be given matrices of size $k\times 1$ and $k\times n$ then $$\mathbf{Y} = \mathbf{a} + \mathbf{b}^T \mathbf{X}$$ is a random ...

2

Here, it means the normal PDF: $$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$ The $\mu,\sigma^2$ in given side means that you can treat them as known quantities.

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

$p_\theta(\bf x)$ is the constructed likelihood function(intractable model density) and not marginal likelihood. The chain of partial derivatives indicates that the auxiliary function(CDF) used is differentiable in all dimensions. "Learning in Implicit Generative Models" (Mohamed and Lakshminarayanan, 2016) is a good reference concerning the learning and ...

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 $Y$ are 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 \\ &=...

1

From comment while question was closed: What is possible is a Poisson process with a rate which varies over time. That could give the sort of graph you are seeing on Google. As you commented, this would be a nonhomogeneous Poisson process If you integrate that rate over a particular time period, then the actual number of arrivals in the whole of the time ...

1

The converse is also true and you can write the joint pmf as the multiplication of the two marginal pmfs.

1

Your PDF doesn't integrate to $1$, so you need a suitable scalar in front of it, e.g. $3/16$ instead of $1/16$. But, let's assume it's valid. You don't need to transform boundaries because the integration is in terms of $x$. The last term is also correct because both your boundaries and the integral $x^2f(x)$ is calculated correctly.

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