New answers tagged

1

You can analyze this graphically with a ROC curve which relates as well to areas under Gaussian curves to the right of some value. You will be looking for the point where the ROC curve intersects the diagonal given by $P(TP) = Y\cdot P(FP)$ (There can be zero, one, or two solutions). You can do this easily computationally with a search algorithm. I doubt ...


2

You computed the CDF by using the proper integral of the PDF $$\int 2x^{-2} dx = \frac{-2}{x} + C $$ But what you forgot is to use the correct integration constant (or use a definite integral). Your CDF is not $$F(x) = \frac{-2}{x}$$ But instead $$F(x) = \begin{cases} 0 &\quad \text{if} \quad x \leq 2 \\ \int_2^x 2u^{-2} du = 2 - \frac{2}{x} &\quad ...


6

The mean of a variable $X$ can be computed as $$\mu_X = \int_{0}^{\infty}1-F(x)dx - \int_{-\infty}^{0} F(x)dx $$ The mean of a shifted variable $X-\mu_X$ (which equals zero) is computed as $$0 = \int_{0}^{\infty}1-F(x+\mu_X)dx - \int_{-\infty}^{0} F(x+\mu_X)dx $$ Or $$ 0 = \int_{\mu_X}^{\infty}1-F(x)dx -\int_{-\infty}^{\mu_X} F(x)dx$$ Which is ...


1

We may draw insight from considering a generalization of truncation. As a point of departure, then, I propose viewing truncation as an extreme example of locally modifying the probability. (This is known as change of measure in the literature on stochastic processes.) That is, in the Wikipedia setting where a distribution $\lambda$ is truncated to the left ...


3

As noted in comments, Wikipedia gives $$x = \Phi^{-1}( \Phi(\alpha) + U\cdot(\Phi(\beta)-\Phi(\alpha)))\sigma + \mu$$ for generating a random variate $x$ from a truncated normal distribution Translating this expression to your question, I suspect you want $$F^{-1}(x; \mu, \sigma,\alpha,\beta) = \Phi^{-1}\left( \Phi(\alpha) + x\cdot(\Phi(\beta)-\Phi(\alpha))\...


1

$$F^{-1}(p; \mu, \sigma,a,b) = \Phi^{-1}\left(\Phi\left(\frac{a-\mu}{\sigma}\right) + p\left(\Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)\right)\right)\sigma + \mu,\quad p \in (0, 1)$$


11

The CDF is defined the probability that your random variable takes a value less than or equal to a real number: $$ F\colon\mathbb{R}\to\mathbb{R}, x\mapsto F(x):=P(X\leq x). $$ And of course if $X$ only has probability mass on the natural numbers, the probability that $X$ is (e.g.) less than or equal to $3$ is the same as the probability that it is less than ...


-1

In practice, the first step would be to generate a random deviate for the first branch of the specified CDF with half of the probability. So, let ½ U (where U is a Uniform random deviate on (0,1] ) = F(x) = ½ ${x^2}$, where x lies between 0 and 1, producing the obvious answer (per the Monte Carlo inversion approach for deriving random deviates) that ${X = ...


5

Here's some intuition. Let's use a discrete example. Say after an exam the students' scores are $X = [10, 50, 60, 90]$. But you want the scores to be more even or uniform. $h(X) = [25, 50, 75, 100]$ looks better. One way to achieve this is to find the percentiles of each student's score. Score $10$ is $25\%$, score $50$ is $50\%$, and so on. Note that the ...


1

As @whuber comments, your CDF is not correct. My plot of your CDF (from R) is shown below. It is OK for a PDF to be discontinuous, but the CDF of a continuous random variable must be continuous. First, you might try differentiating your CDF (piecewise) to see whether the result is your PDF. curve((exp(x)-1)/exp(x), 0,2, xlim=c(0,10), ylim=0:1,ylab="...


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