# Tag Info

14

You does not need an approximation here. Use properties of moment generating functions, $X$ is standard normal so $X^2$ is chisquared with one df, with moment generating function $M_{X^2}(t)=\frac1{\sqrt{1-2t}}$ (for $t<1/2$.) Then note that $$\DeclareMathOperator{\E}{\mathbb{E}}M_X(t)=\E e^{t X}$$ is the definition, so that \E e^{-X^2}=M_{X²}(-1)=\... 8 There's no need to "approximate" when you can derive the exact value of \mathbb{E}[f(X)] . 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)~dx\\ &= 2\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} ... 6 The straightforward extension of the univariate case \mathbb{E}\big[X\big] = \int_{\mathbb{R}}xf(x)dx $$to the bivariate one is$$ \int_{\mathbb{R}\times\mathbb{R}}\color{blue}{(x_1,x_2)}f(x_1,x_2)d(x_1,x_2) $$rather than$$ \int_{\mathbb{R}\times\mathbb{R}}\color{blue}{x_1x_2}f(x_1,x_2)d(x_1,x_2). $$While the notation might be unusual, it can be ... 4 If the listed RVs are independent, their covariances should be 0. So,$$\operatorname{cov}(X,Y-X)=-\operatorname{var}(X)+\operatorname{cov}(Y,X)=0\operatorname{cov}(X,Z-Y)=\operatorname{cov}(X,Z)-\operatorname{cov}(X,Y)=0$$From the first equation, \operatorname{cov}(X,Y)=\operatorname{var}(X)=1. Substituting this into the second equation gives \... 4 If your random variable is bivariate, then every realization is a pair of numbers. The expectation of a random number can be thought of as "the long-run average". A long-run average of a large number of pairs makes most sense as a pair of numbers, not a single number. Specifically, as the pair of the separate long-run averages. Which is why the ... 4 Since this looks like a homework question, I'll give some hints instead of a complete solution. Assume independent and identical trials (o/w the question should provide more information). Let X be defined as the number of trials until success, with success probability p. X is a well-known RV, i.e. geometric. You're given the expected value, which is ... 3 Assuming the distribution is valid, for which you need to clarify its support, you've already solved it:$$\mathbb E[X+Y^2|Y]=\mathbb E[X|Y]+\mathbb E[Y^2|Y]=1+1/Y+Y^2$$Because given Y, expected value of Y^2 is itself. 3 If X is a bivariate random variable [X_1,X_2], then X_1 and X_2 are functions of X (projections of X onto the individual components). Thus, E[X] is the vector \big[E[X_1],E[X_2]\big] while what you are computing with the double integral that you wrote is E[X_1X_2]. 2 Let's offer a "dissenting" view: Ratios and inverses of random variables can be fine in the following sense: It may be the case that in many cases they do not possess moments But it is also the case that in many cases they result in recognizable, "named" and exhaustively studied distributions. ...and there is distribution-life beyond ... 2 Following from the third line, E[Y_i-\bar Y|\mathbf X]=(\beta_0+\beta_1X_i)-(\beta+\beta_1\bar X)=\beta_1(X_i-\bar X). When substituted that back, we have$$\begin{align}E[\hat \beta_1|\mathbf X]&=\sum_{i} \beta_1g_i(\mathbf X)(X_i-\bar X)=\beta_1\sum_i\frac{(X_i-\bar X)}{\sum_j (X_j-\bar X)^2}(X_i-\bar X)\\&=\beta_1\frac{\sum_i (X_i-\bar X)^2}{\...

2

Both notations are correct and you can do them. Once you conditioned on a random variable or vector, you can treat any function of it that doesn't include other RVs as a constant and take it out from the expectation.

2

The probability of dying at year i where i is 1 to 50 is $$p(i) = 0.015 \cdot 0.985^{(i-1)}$$ the remaining probability, say p(51), is the probability of dying from cause Z $$p(51) = 0.985^{50}$$ Furthermore dying at year i removes (51-i) years from your life And the total loss is the sum of the products of the probabilities of death at year i and the ...

2

These negative moments of random variables are in general difficult to obtain closed-form expressions for. You already have $\text{E}[X^{-1}]$ in the form of an infinite series, but what you ultimately want is a representation as a finite sum or product. As I learned from my mentor, sometimes it is possible to get this using the generating function of $X$ as ...

2

YES. You have given that $Y \mid X=x \sim \mathcal{N}(0,1), ~~\text{for all$x$within the range of$X$.}$ This implies that the random variables $X$ and $Y$ are independent, and the conclusion follows. A more intuitive answer: You have given a distribution of $Y$ conditional on some $X$, but the condition is not used at all when stating the distribution. ...

1

A useful approach to debugging a string of equalities is an example or two, so you can check where the equality stops holding. The simplest example I can think of for this is $Y$ being a constant that isn't 0, 1 or -1. So, let $Y=\mu_Y$ be a positive constant that isn't 1, and $\sigma^2_Y=0$. The first three equalities are just expanding definitions, so the ...

1

One possibility is $$Z=X\left(1-\frac{y}{\mathbb E[X]}\right)$$ In that case you will have $$\mathbb E[Z] = \mathbb E[X]-y$$ and, if $\mathbb P(0 \le X \le 1.8)=1$ and $0 \lt y \le \mathbb E[X]$, then $\mathbb P(0 \le Z \le 1.8)=1$

1

I expect $X$ and $Y$ are strictly positive random variables. Then using indpendence, Jensen inequality and identical distributions it follows $$\mathbb{E}\left[\frac{X}{Y}\right]=\mathbb{E}[X]\mathbb{E}\left[\frac{1}{Y}\right]>\mathbb{E}[X]\frac{1}{\mathbb{E}[Y]}=\frac{\mathbb{E}[X]}{\mathbb{E}[X]}=1.$$

1

You can see it as $$\hat{\beta}=(X^tX)^{-1}X^ty = (X^tX)^{-1}X^t(X\beta+\varepsilon) =(X^tX)^{-1}X^tX\beta + (X^tX)^{-1}X^t\varepsilon$$ So you have that $$\hat{\beta}=\beta + (X^tX)^{-1}X^t\varepsilon$$ And then, it is straightforward to see that  \mathbb{E}(\hat\beta) = \mathbb{E}(\beta + (X^tX)^{-1}X^t\varepsilon) = \beta + (X^tX)^{-1}X^t\mathbb{E}(\...

Only top voted, non community-wiki answers of a minimum length are eligible