# Tag Info

8

I'll try to explain it in linear case. Consider the linear model $$Y_i=\sum_{j=1}^{p} \beta_jX_{i}^{(j)}+\epsilon_i, i=1,...,n.$$ When $p \leq n$ (number of independent variables less or equal then number of observation) and design matrix has full rank, the least squared estimator of $b$ is $$\hat{b}=(X^TX)^{-1}X^TY$$ and prediction error is $$\dfrac{\| X(\... 8 Assuming independent X_i, the mean \frac{1}{n}\sum X_i is also normal, i.e. N(0,1/n). Absolute value of it is Half-normal, which has mean E[Y_1]=\frac{\sigma\sqrt{2}}{\sqrt{\pi}}=\sqrt{\frac{2}{n\pi}}. For Y_2 we can find the expected value directly:$$E[Y_2]=\frac{1}{n}\sum_{i=1}^n E[|X_i|]=E[|X_i|]=\sqrt\frac{2}{\pi}$$This means \sqrt n E[Y_1]=... 6 Let's try the usual preliminaries: simplify by choosing appropriate units of measurement and exploiting the symmetry assumption. Reframing the question Change the units of X so that its mean is m=1: this will not alter the truth of the inequality. Thus the distribution F of X is symmetric about 1 and the range of X is within the interval [0,2]... 6 For a single covariance you only need the bottom equation. The bounds are that the covariance cannot be greater than the product of the standard deviations (and cannot be less than the negative of the same value). However for a covariance matrix of more than 2 terms there is an additional limit, the matrix has to be positive semi-definite (or positive ... 6 Answer: Whatever the distribution of X_1,...,X_n,$$\mathbb{E} Y_2 \geq \mathbb{E} Y_1.$$Details: For any n numbers X_1,..., X_n it is true that$$ \sum_i |X_i| \geq |\sum_i X_i|$$and dividing both sides by n:$$ \frac{1}{n}\sum_i |X_i| \geq \frac{1}{n}|\sum_i X_i| = |\frac{1}{n}\sum_i X_i|.$$Now, the key word is 'any', that means that ... 5 Remark: the sign seems to be flipped. Here is how we can prove if X> 0, then$$E(X \ln X) \ge E(X) E(\ln X)$$We can apply Jensen's inequality twice. x\ln x is convex, hence we have,$$E[X] \ln (E[X]) \le E(X\ln X)$$\ln x is concave, hence we have$$\ln (E(X)) \ge E(\ln X).$$Combining the two inequalities,$$E(X \ln X) \ge E[X] \ln(E(X)) ...

5

You automatically get $||X + Y|| \le ||X|| + ||Y||$ by Minowski's inequality. The last part is simple arithmetic showing that: $\sqrt{A + B} \le \sqrt{A} + \sqrt{B}$ when $A, B > 0$. which is a simple direct proof. In fact, I think the whole thing is a direct consequence of Minowski's inequality.

5

One way to do it is to proceed from the answer you linked in comments, at the second last line (at this point no properties of the normal have been used). Note that $Y$ is just a standardized $X$; i.e. $X=\mu+\sigma Y$: \begin{align}\operatorname{Cov}(X,X^2)&=\sigma^3\operatorname{Cov}(Y,Y^2)+2\mu\sigma^2\operatorname{Cov}(Y,Y)\\ &=\sigma^3\gamma_1+...

4

\begin{align} EX &= \int x\cdot \mathbf{1}_{x< a} + x\cdot\mathbf{1}_{x\geq a} \ dP \\ &= \int x\cdot \mathbf{1}_{x< a}\ dP\\ &< \int a\cdot \mathbf{1}_{x< a}\ dP \\ &= a, \end{align} so yes.

4

Here is an answer (to a related question) which provides a valid generalization of $\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$. What is the relation between $\mathbb{E}X^r$ and $\mathbb{[E}X]^r$, for all possible values of $r$, when $X$ is a positive random variable? This can be answered by applying Jensens's inequality, based on the ...

4

First, @Dilip Sarwate comments, for such ratio variables mean and variance often do not exist and then there is little to expect. For a detailed discussion of this see I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. But, let us assume a case where expectation and variance exists. ...

4

I will focus on the case where $|\rho| < 1$. We are given that $\mathbb{E}[X_1] = \mathbb{E}[X_2]=0$, $\operatorname{Var}[X_1]=\operatorname{Var}[X_2]=1,$ and the correlation is $\rho$. From there we can deduce that $\mathbb{E}[X_1^2]= \mathbb{E}[X_2^2]=1$ and $\mathbb{E}[X_1X_2]=\rho$. Let $g(x_1, x_2; t)=x_1^2+2tx_1x_2+x_2^2$ where $|t| < 1$. We ...

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2

$$E|X|^{r-1} = E\left[|X|^{r-1} \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r-1} \mathbb{1}_{(|X|> 1)} \right]$$ $$\leq E\left[1 \times \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r-1} \mathbb{1}_{(|X|> 1)} \right]$$ $$\leq E\left[1 \times \mathbb{1}_{(|X|\leq 1)} \right]+E\left[|X|^{r} \mathbb{1}_{(|X|> 1)} \right]$$ $$\leq 1 + E|X|^r$$ Where ...

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