# Tag Info

3

This is not related at all. For example let $X_{i}$ be i.i.d Gaussian distributions of pdf $$\frac{1}{\sqrt{2\pi}}{e^{-x^2/2}}$$ Then by definition we have $Cov(X_{i},X_{j})=0$ because $X_{i},X_{j}$ are independent. On the other hand if we have $$X\sim N(0,1), X_{1}\sim aX+b, a,b\not=0$$ Then we have $$Cov(X,X_{1})=a^2\not=0$$ Thus $X,X_{1}$ are ...

3

Just normalized X to the N(0,1) and used a Z table! $$P(X<7)=P(X-3/\sqrt4<7-3/\sqrt4)=P(Z< \frac{7-3}{\sqrt4})$$ Just use a Z table now for this number.Same for the others, let me know if this helped.

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You are partially correct. You need to integrate the PDF so you get the CDF, then it's: $$P(X<7)=F(7)$$and$$P(X≥7)=1-F(7)$$ Note: $F$ here is the CDF. Normally a lowercase $f$ indicates the PDF. As you have a normal distribution though, you can get around integrating that expression, and instead use the table for the normalized version as Steven Ramos ...

3

If $f(x)$ is the probability density function, then $P(X<X_0)$ is an integral of $f(x)$: $$P(X<X_0) = \int_{-\infty}^{X_0} f(x)\,dx$$ But for the probability density function $f$ for the normal distribution this integral is difficult to express in terms of commonly used functions. It is usually tabulated or calculated numerically. For the other ...

3

Such $k$ does not exist. We have that \begin{align*} |X+k|-|X-k| &=\frac{(|X+k|-|X-k|)(|X+k|+|X-k|)}{|X+k|+|X-k|}\\ &=\frac{|X+k|^2-|X-k|^2}{|X+k|+|X-k|}\\ &=\frac{4kX}{|X+k|+|X-k|}. \end{align*} Hence, $|X+k|-|X-k|>0$ if and only if $X>0$ when $k>0$ and $|X+k|-|X-k|>0$ if and only if $X<0$ when $k<0$. The probability does not ...

2

Notice that $\ln(\color{blue}{\sqrt{\color{black}{x}}}) = \ln(x^{\color{blue}{\frac{1}{2}}}) = \color{blue}{\frac{1}{2}}\ln(x)$ for all $x > 0$. Using this identity, let us re-write the maximum entropy, $\frac{1}{2} + \ln(\sqrt{2\pi}\sigma)$, as follows: \begin{align} \frac{1}{2} + \ln(\sqrt{2\pi}\sigma) &= \frac{1}{2} + ... 2 For continuous distribution like Normal/Gaussian we compute the differential entropy. You can find the derivation here http://www.biopsychology.org/norwich/isp/chap8.pdf For more info on differential entropy I recommend the book "Elements of Information Theory" by Cover and Thomas. 2 Hint: If k>0 then:|x-k|<|x+k|\iff x>0$$In words: to be more close to k>0 instead of -k (on the same distance of 0 as k, but on the other side) it is necessary and sufficient that x>0. edit: Consequently P(|X-k|<|X+k|)=P(X>0)=0.5\neq0.7. Conclusion: k cannot be positive. Likewise it can be shown that k cannot ... 2 If X\sim N(\mu,\sigma), then Y=\frac{X-\mu}{\sigma}\sim N(0,1)  and$$ \mathbb{P}[\mu-k\sigma \leq X \leq \mu+k\sigma] = \mathbb{P}[-k\leq Y \leq k].$$Can you recognize \Phi in the RHS now? 1 Heavens, no. Recall that for two functions f,g\in L^2(\mathbb{R}^n), their inner product is \langle f,g\rangle = \int fg\ dx. In fact, two positive functions whose supports intersect can never be orthogonal. This is because their product is positive, hence its integral is positive. As Gaussians are supported on all of \mathbb{R}^n and are positive, ... 1 The distribution of \|X\|^2/\sigma^2 is \chi^2_k. Since for k=1$$ P(|x| \le 3 \sigma) = .9973002 $$I interpret this as the question for which r in the case of k = 3$$ P(\|X\| \le r \sigma) = .9973002. $$This is r = \sqrt{F^{-1}(.997302)} = 3.76205 where F is the cumulative distribution function of a \chi^2_3 random variable. R command: ... 1 You have already gotten some good answers, I thought I could add something more of use which is not really an answer, but maybe good if you find differential entropy to be a strange concept. Since we can not store a real or continuous number exactly, entropy for continuous distributions conceptually mean something different than entropy for discrete ... 1 Given X - a continuous random variable (eg. \mathcal{N}(0,1)), we have \mathbb{E}\left[f(X)\right] = \int_{-\infty}^{+\infty}f(x)g(x)dx, where g(x) is the probability \Bbb{density} function (pdf) of a \mathcal{N}(0,1) variable and f is a measurable function of X. Since you mention the cumulative density function (cdf), I will refer to that ... 1 We have to show if X\sim \mathcal N(\mu, \sigma^2) and Z=\frac{X-\mu}{\sigma}\sim \mathcal N(0,1) then P(X\leq w)=P(Z\leq \frac{w-\mu}{\sigma}). Z=\frac{X-\mu}{\sigma}\Rightarrow Z\cdot \sigma+\mu=X P(X\leq w)=P(Z\cdot \sigma+\mu\leq w)=P(Z\cdot \sigma\leq w-\mu)=P(Z\leq \frac{w-\mu}{\sigma}). 1 (a) Yes, the sum of n iid Geometric Distributed Random Variables has a Negative Binomial distribution, and that is the right moment generating function for the given one. (b) is okay, and see also (d) below. (c) Well, \mathsf e^{4t/3} is the moment generating function for a Degenerate Distribution. In this case ... 1 In linear regression with Gaussian (and heteroscedastic) noise, our model assumes that for n observations of data, for each i \in [n],$$Y_i = \beta X_i + \epsilon_i,$$where \epsilon_i is our ERROR term for the ith observation (note that residual e_i is an estimator of \epsilon_i) Such that \epsilon_i \sim N(0,\sigma^2_i). NID means ... 1 To find expressions like P(X\geq x) where X has normal distribution observe that$$P(X\geq x)=P(\sigma U+\mu\geq x)=P\left(U\geq\frac{x-\mu}{\sigma}\right)$$where U has standard normal distribution. So actually for z=\frac{x-\mu}{\sigma}:$$P(X\geq x)=1-\Phi\left(z\right)$$You can find this z by substituting. From here tables come in. Also ... 1 For every x\in\mathbb R:$$\Phi\left(-x\right)=1-\Phi\left(x\right)$$Substituting x=z_a leads to$$\Phi\left(-z_{a}\right)=1-\Phi\left(z_a\right)=1-a=\Phi\left(z_{1-a}\right)$$Conclusion:$$-z_a=z_{1-a}$$Or equivalently (but a bit nicer):$$z_a+z_{1-a}=0$$1 I think it must be proved that \mu=\mathcal N(0,\sigma^2) but for convenience I will also preassume that \sigma=1 If \phi denotes the characteristic function then:$$\phi(t)=\phi\left(\frac{t}{\sqrt2}\right)^2 Note that this can be repeated to arrive at $\phi(t)=\phi(\frac{t}2)^4$ and can be repeated again. Actually with this it can be shown that ...

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