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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.


3

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 $i$th 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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