New answers tagged

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 ...


0

If you by a time series means a concrete series of nmbers/observations attached to some time points, then it is neither stationary or not stationary, neither does it have an expectation (or not!) Those concepts apply only to time series models, that is, some probability model for a time series. So by just looking at your blue wawy curve, nobody can answer ...


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 ...


0

Clearly, dividing the top and bottom of the fraction by the non-zero numerator implies a value for the expectation to be bounded between 0 and 1.


0

If $Z \sim \mathsf{Norm}(0,1)$ and, independently, $Y\sim \mathsf{Chisq}(5),$ then (by definition of the t distribution) $W^\prime = \frac{Z}{\sqrt{Y/5}} = \sqrt{5}\frac{Z}{\sqrt{Y}} = \sqrt{5}W \sim \mathsf{T}(5),$ which has $Var(W^\prime) = 5/(5-2) = 5/3.$ Thus $Var(W) = \frac{1}{5}\cdot\frac{5}{3} = \frac{1}{3}.$ Generalize. Simulation in R: With a ...


0

It is probably not easy to exactly compute the distribution. For instance, when $\theta = 1$ then $\bar X$ follows the Bates distribution. That is not easy to integrate over. However, we can approximate the mean by using a simulation or by using a function that approximates the distribution. Using an approximation of the distribution We can approximate the ...


0

When writing (correctly) $$\mathbb E[X]=\int_{0}^{1} \int_{x}^{1} 2x \,\mathrm dy \,\mathrm dx$$ one skips intermediate steps: \begin{align} \mathbb E[X] &= \int_{\mathbb R^2} x f(x,y)\,\,\mathrm d(x,y)\tag{definition}\\ &= \int_{\mathbb R^2} x\times2\times\mathbb I_{\underbrace{\{(x,y);\,0\le x\le y \le 1\}}_{\text{support set}}}(x,y) \,\,\mathrm d(...


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.


-2

well to be exact... $$Y|X=x_0 \sim \mathcal{N}(0,1)$$ i.e.: $$Y|X=-7 \sim \mathcal{N}(0,1)$$ $\require{enclose}$ you can't have a new RV from this $\enclose{horizontalstrike}{Z=(Y|X)}$ because it require some input, but from $Z=(Y|X=x_0)$ so does: $Y^2|X=-7 \sim \chi^2_1$? yes although $Y^2|X=-6 \sim \chi^2_1$? this you can't reason from given assumptions (...


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.


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 ...


1

I expect $X$ and $Y$ are strictly positive random variables. Then using indpendence, Jensen inequality and identical distributions it follows \begin{equation} \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. \end{equation}


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}} ...


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)=\...


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 $\mathbb P(0 \le Z \le 1.8)=1$


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 $\...


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}(\...


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.


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 ...


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 $E[X_1X_2]$.


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 ...


0

No. You also need additional information to obtain $\mathrm{E}(A)$. In particular, the probability distribution of $B$ helps. We can write $$ \mathrm{E}(A) = \sum_b \mathrm{E}(A|B=b) \mathrm{P}(B=b) $$ This means that if you know $\mathrm{E}(A|B)$ and $\mathrm{P}(B)$, then you can calculate $\mathrm{E}(A)$.


3

If $(X,Y)$ is a random vector, then the mean of the random vector is defined as $(E[X], E[Y])$, so you just need to take the mean of each component separately. If $(X,Y)$ has a joint density $f$, then for instance $$E[X] = \int_{-\infty}^\infty\int_{-\infty}^\infty x f(x,y) \, dx \, dy.$$ Note that in cases where $X$ and $Y$ are independent, the joint ...


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