Search Results

For a continuous random variable, the raw $k$-th moment is by LOTUS: \begin{align}\large \color{red}{\mathbb{E}\left[{X^k}\right]} &= \displaystyle\int_{-\infty}^{\infty}\color{blue}{X^k}\,\,\color{green …

$\newcommand{\y}{\mathbf y}$We have $$ E_{\theta|\y}\left[f(\theta)\right] = \int f(\theta) p(\theta | \y)\,\text d\theta $$ just by definition of expectation (and you could cite the LOTUS as well), and …

The general calculation for both quantities can be obtained by the application of LOTUS. …

To apply LOTUS, I would need the pdf of $Y_n$. Of course the pdf of the sum of two independent random variables is the convolution of their pdfs. …

Let us apply the Law of the Unconscious Statistician (LoTUS) to obtain : \begin{align*} \mathbb{E}[f(X)] &= \int_{-\infty}^{+\infty} e^{-x^2} \cdot \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right) …

X}\right] && \text{definition} \\ &= \int_{- \infty}^{\infty} x \cdot p_X(x) dx&& \text{just definition of expectation} \\ &= \int_{- \infty}^{\infty} e^{t x} \cdot \lambda e^{-\lambda x} dx&& \text{LOTUS …

car_model %in% c('Lotus Europa', 'Ford Pantera L', 'Ferrari Dino', 'Maserati Bora', 'Volvo 142E')) model = lm(mpg ~ disp * am, data = df) summary(model) #> #> Call: #> lm(formula = mpg ~ disp * am, data …

The LOTUS now involves two-dimensional real integrals, frequently expressed as Riemann or Lebesgue integrals. … .$$ LOTUS asserts this can be written as an "ordinary" integral, $$\mathbb{E}(Z) = \int_\Omega z\,p(z)dz.$$ The (bivariate) LOTUS, applied to $Z$ in in terms of $(X,Y)$, says this can be written $$\ …

1-y, & 0<y<1 \end{cases}$$ Here is the plot: In this way, $$\text{Var}\left[(U-U')^2\right]=\text{Var}[Y^2]= \mathbb E\left[Y^4\right]-\left[\mathbb E\left[Y^2\right]\right]^2\tag 1$$ Applying LOTUS …

I wouldn't know of an R implementation of LOTUS. … The binaries for the original LOTUS implementation are available from Kin-Yee Chan's web page (http://www.stat.nus.edu.sg/~kinyee/lotus.html) but I think Wei-Yin Loh's recommendation nowadays is to use …

You can do this directly using the distribution of the previous step and LOTUS or by first finding the distribution of $Z$. Hint: the gamma family is a truly large family of distributions. …

Specifically, I am looking for a logistic regression tree implementation in R based on a specific algorithm (LOTUS - Logistic regression Tree with Unbiased Splits) developed by Kin-Yee Chan and Wei-Yin …

I worked it out and found how $\sum_i -\ln X_i = \sum Y_i \stackrel{\text{d}}{=} \Gamma(n,\theta)$ (through the hints supplied byJohnK), then I immedialty used LOTUS and found $E[\hat \theta] = \dfrac{ …

Im guessing for the mean we could use LOTUS such that $$ \mathrm{E}[Y] = \int_{0}^{\infty}\dots \int_{0}^{\infty} \ln{\left(\sum_{i=1}^{n} X_{i}^{2}\right)} \prod_{i=1}^{n} P(X_i|\sigma) \; \mathrm{d}X …

cdots + g(99)P(X=99) + \cdots $$ which, if we are so inclined, we can recombine into gobbledygook involving $\displaystyle \sum$ by writing $$E[Y] = E[g(X)] = \sum_i g(x_i)P(X=x_i)$$ which looks like LOTUS …