Search Results

Solving for $a$,$$a=\pm\frac1{\sqrt{V(X)}}$$ then substitute to the first condition that you found. $$b = -aE[X]=\mp \frac{E[X]}{\sqrt{V(X)}}$$ I think the purpose of the exerise is to construct tha …

Let $s \in [0,2]$ \begin{align} f_S(s) &= \int_{-\infty}^\infty f_{U_1}(t)f_{U_2}(s-t) \, dt \\ &= \int_0^1 1\cdot f_{U_2}(s-t) \, dt \end{align} Now we just need to integrate over the region when $ …

Guide: Part $a$ looks fine. Part $a$ is describing the event that at least $1$ will not watch the movie. Its complement is both will watch the movie which is not part $b$ is asking for. For part $b$ …

You are done. $$P(X=x) = \begin{cases}\frac{3}{5}, & ,x=1 \\ \frac{2}{5}\cdot\frac{3}{4}, & ,x=2\\ \frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{3} & ,x=3\\ 0, & \text{,Otherwise}\end{cases}$$

Assuming uniform splitting, the expected longer part can be described as \begin{align*} \mathbb{E}[\max(X,1-X)]&=\int_0^1 max(x,1-x) dx\\ &=\int_0^{0.5} (1-x) dx + \int_{0.5}^{1} x dx \\ &= 0.5-\fra …

By Chebyshev's inequality, $$P \left(|\hat{\alpha_n}-\alpha| >\epsilon\right)\leq \frac{v(\hat{\alpha_n})}{\epsilon^2}$$ $$\lim_{n \to \infty}P \left(|\hat{\alpha_n}-\alpha| >\epsilon\right)\leq \lim …

Let the smallest element be $m$ and let the sum be $M$. We want to map $m$ to $0.5$ and $M$ to $1$. View it as you are trying to interpolate $(m, 0.5)$ and $(M, 1)$. We have $$\frac{y-1}{x-M}=\frac{0. …

The pdf of $X$ can be computed as such: $$f_X(x; B) = \int_b f_X(x; b) f_B(b) \, db$$ Here are some readings on marginal distribution.

Given $x$, we compare $P(D_1|x)$ and $P(D_2|x)$ and decide which has a higher value. To compute these values though, some assumptions and prior knowledge are required. From Bayes' rule, this is equiva …

A loss function $$l(y, f)$$ measures how good we can use $f$ to predict $y$. The smaller the result of the loss function $l$ is, the better the function $f$ is in performing the prediction task. Ex …

$$Pr(Y \leq y) = Pr(X_1 \leq y, \ldots, X_n \leq y) = \prod_{i=1}^n Pr(X_i \leq y)$$ $$Pr(Z \leq z) = 1 - Pr(Z> z) =1-Pr(X_1 > z, \ldots, X_n > z) = 1-\prod_{i=1}^n Pr(X_i >z )$$

Guide: You have listed out all the possibilities. Each of the possibilities happen with probability $\frac16$. For each of the possibility, compute their sum, for example the sum for $(1,2)$ would b …

As you did not specify the objective function, it is hard to justify which grouping is better. One interpretation of your task is that this could be a clustering problem. A famous algorithm would b …

Since it is possible to have a slope that is greater than $1$, it is not true. In terms of simple linear regression, the slope is given by $\frac{Cov(X,Y)}{Var(X)}=r_{xy}\frac{s_y}{s_x}$ which is not …

Here, the rate refers to the expected number of failure per hour. $\lambda$ can be any positive number.