If $X\sim  N(0, \sigma^2)$, how can I compute $\operatorname{Var}(X^2)$? Here is my idea... but I cannot get there. $$\operatorname{Var}(X^2) = E(X^4) - (E(X^2))^2$$
121974
What is the mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$?  It is interesting to note that Gaussian R.V here is zero-mean and non-central Chi-square Distribution doesn't work.  Thanks.
58775
Let $X_1, \ldots, X_n$ be i.i.d. random variables with expectation $a$ and variance $\sigma^2$, taking only positive values. Let $m &lt; n$. Find the expectiation of $\displaystyle\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$.   My attemps to solve this probles are rather straightforward. Denote $X = X_1 + \cdots + X_m$ and $Y = X_{m+1} + \dots + X_n$. So, $X$ has the expectation $ma$ and the variance $m\sigma^2$. And $Y$ has the expectation $(n-m)a$ and variance $(n-m)\sigma^2$. And also $X$ and $Y$ are independent. So we can compute the expectation by the definition $\mathbb{E}\displaystyle\frac{X}{X+Y} = \int\limits_{\Omega^2}\frac{X(\omega_1)}{X(\omega_1) + Y(\omega_2)}\mathbb{P}(d\omega_1)\mathbb{P}(d\omega_2)$. But we do not know the distribution, so we do not have chance to calculate it.  I would be glad to any help or ideas!
88708
Let there be a random variable $X$ and $N(0, \lambda ^2)$ given. I am to find $E[X^n]$ knowing that $\Gamma \big(\frac{1}{2} \big) = \frac{\pi}{2}$.     I started with the definition: $$E[X^n] = \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I figured out that: $$E[X^n] = \begin{cases} 0 &amp;\text{for } n = 2k + 1; k \in \mathbb{N} \cup\{0\}\\\text{something}&amp;\text{for } n = 2k; k \in \mathbb{N} \end{cases}$$    So now: $$ \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x =  \frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I substitute: $$t = \frac{x}{\lambda}$$ $$\mbox{d}x = \lambda \mbox{d}t$$  Thus: $$\frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x = \frac{2 \lambda^{n}}{\sqrt{2 \pi}}\int \limits_{0}^{\infty}t^ne^{\frac{-t^2}{2}} \mbox{d}t.$$ By integrating the formula above by parts I do get:  $$\frac{2 \lambda^{n}}{\sqrt{2 \pi}}(n-1) \int \limits_{0}^{\infty}t^{n-2}e^{\frac{-t^2}{2}} \mbox{d}t$$ I don't really know where to go from here. And unfortunately I don't see how can I use the Gamma Function, which is given by this formula: $$\Gamma (z) = \int \limits_{0}^{\infty} x^{z-1}e^{-x} \mbox{d}x,$$ isn't it?
71819
This is a more general treatment of the issue posed by  . After deriving the asymptotic distribution of the sample variance, we can apply the Delta method to arrive at the corresponding distribution for the standard deviation.    Let a sample of size $n$ of i.i.d. non-normal random variables $\{X_i\},\;\; i=1,...,n$, with mean $\mu$ and variance $\sigma^2$. Set the sample mean and the sample variance as  $$\bar x = \frac 1n \sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1} \sum_{i=1}^n(X_i-\bar x)^2$$  We know that  $$E(s^2) = \sigma^2, \;\;\; \operatorname {Var}(s^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right)$$  where $\mu_4 = E(X_i -\mu)^4$, and we restrict our attention to distributions for which what moments need to exist and be finite, do exist and are finite.  Does it hold that  $$\sqrt n(s^2 - \sigma^2) \rightarrow_d N\left(0,\mu_4 - \sigma^4\right)\;\; ?$$
61786
Same as . Let $X$ be a square integrable random variable on $(\Omega,\mathcal{F},P)$. Let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. Define the conditional variance of $X$ given $\mathcal{G}$ by $$\operatorname{var}(X\mid\mathcal{G})=\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}]$$ Prove the formula  $$\operatorname{var}(X)=\operatorname{var}(\Bbb{E}[X\mid\mathcal{G}])+\Bbb{E}[\operatorname{var}(X\mid\mathcal{G})]$$  1 I don't understand part of the presented solution.  2 Is my attempt also correct?  3 Alternate solutions?  1 I don't understand this part of the solution wherein  $$\operatorname{var}(X\mid \mathcal{G})=\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2 \mid \mathcal{G}]$$  $ =\mathbb E[ X^2 \mid \mathcal G] - \mathbb E [X \mid \mathcal G]^2 $. I think it is making use of the fact that $\mathbb E [X \mid \mathcal G]$ is $\mathcal G$-measurable in saying:  $\mathbb E[XE[X\mid\mathcal G]\mid\mathcal G] = \mathbb E[X\mid\mathcal G] \ E[X\mid\mathcal G] = E[X\mid\mathcal G]^2$, but isn't $E[X\mid\mathcal G]$ supposed to be bounded? How to show this?  My attempt:  $E(X^2) &lt; \infty$  $\to E(X) &lt; \infty$  $\to E(X\mid\mathcal G) &lt; \infty$, but I don't think .  2 My own attempt: I tried evaluating $$E(\operatorname{var}(X\mid\mathcal{G}))=E(\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}])$$  Is it correct to say that $E(\Bbb{E}[(X-\Bbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}]) = E([(X-\Bbb{E}[X\mid\mathcal{G}])^2])$?  After expanding the square I encounter $E[XE[X\mid\mathcal G]]$ and rewrite  $= E[E[XE[X\mid\mathcal G]]\mid\mathcal G]$  $ = E[E[X\mid\mathcal G] \ E[X\mid\mathcal G]]$ (*)  $ = E[E[X\mid\mathcal G]^2]$  (*) Assuming this is all correct, I am still unsure about this part. Is it really bounded?  3 Is there a way to go about this without saying $E[X\mid\mathcal G]$ is bounded?  4 Is this wrong? It's from Stochastic Calculus class notes, though the problem is for an Advanced Probability class (and the notes don't say anything about bounded weirdly).
96308
I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) &gt; Var(X)$ where $i &gt; 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.  Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?
45342
I wish to use the  to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.  Could you explain how this is possible, e.g. with an example?
80900
I'm working through an econometrics textbook and it's proving that   $$ \sigma^2 = E(\hat{\sigma}^2) = \frac{SSR}{n-2} $$  I followed the proof (an example of which is shown on ) until it reaches this step:  \begin{align} E\left[\displaystyle\sum\limits_{i=1}^n \hat{u}_{i}^{2}\right] &amp;= E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2}\right] + (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] \\ &amp;+ 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] \\ &amp;= \dots \end{align}  The book says that:   The expected value of the first term should be $(n-1)\sigma^2$ The expected value of the second term should be $\sigma^2$. The expected value of the third term should be $2 \sigma^2$ because the term itself can be written as $2(\hat{\beta}_1-\beta_1)^2 \hat{\sigma}_x^2$.   For the first term, I can work backwards and get that:  \begin{align} \hat{\sigma}^2 &amp;= \frac{1}{n-1} \displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2} \\ &amp;= \sigma^2 \end{align}  but it seems like this only works if we assume that $\hat{\sigma}^2$ is an unbiased estimator.  For the second term:  \begin{align} (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} E\left[(n-1) \hat{\sigma}_x^2\right] \\  &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) E\left[\hat{\sigma}_x^2\right] \\ &amp;= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) \hat{\sigma}_x^2 \end{align}  but this doesn't match what the text shows, so I'm a bit confused as to the final step for this term.   For the third term: \begin{align} 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] &amp;= 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n u_{i}(x_{i}-\bar{x})\right] \\ &amp;= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}(x_{i}-\bar{x})\right] \\ &amp;= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}\right]E\left[(x_{i}-\bar{x})\right] \\ \end{align}  I would have thought that because $E[u_i] = 0$, this term would cancel out, so I'm not sure how to derive $2\sigma^2$ from it.  Can someone give me a hint in the right direction for these steps? I'm firmly trying to work through all of the proofs myself with only a little guidance from the book, but I'm not quite there yet.  I noticed , which links to , so maybe with more reading I'll see where to apply that. (If that isn't it, hints are welcome, but I'll still work through those questions to see if it helps)
107370
I would like to know how to show that the functions  $$r_n(t)=\operatorname{sgn}\big(\sin(2^n \pi t)\big)$$ (where $\operatorname{sgn}$ is the sign function) form an orthonormal system but not an orthonormal basis from $L_2([0,1])$.    Progress: I pick two different variables $i,k$ and show that $\langle r_i (t),r_k (t)\rangle =0$, so I can take the integral $\int_0^1  r_i (t) r_k (t)\,dt$, but I do not know how to show that this is 0.
25527
If $X\sim N(0,1)$, integrate to find the moment generating function of a random variable $X^2$ and identify the distribution of $X^2$ using the moment generating function. $$E\left[e^{tX^2}\right]=\int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}\over{\sqrt{2\pi}}}dx$$ which reduces to $$=\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}}dx =\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{x^2(t-\frac12)}}dx$$ and thus I am stuck. I'm sure there must be some trick to this (like completely the square for the mgf of a standard normal variable X) but I can't figure out what it might be. Any hints?
89896
The random variable $X$ is exponentially distributed with parameter $\lambda = \dfrac{1}{2}$. Also, define $Y = -2X + 1$. Calculate $E(Y)$ and $Var(Y)$.  So we have   $$f_X(x) = \dfrac{1}{2} e^{-\dfrac{1}{2}x} $$  if $x\ge0$ and $0$ otherwise.  But how to calculate $E(Y)$ and $Var(Y)$ ?
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
22712
... and why ?  Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:    $E(X_1-X_2)=\mu_1-\mu_2$ $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$   Now let's say $X_1$, $X_2$  are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?   This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.
771
Let $X_1, X_2, \ldots X_n$ be random variables with mean $\mu$ and variance $\sigma^2$. We define the sample mean $\bar{X} = \sum_{i = 1}^n X_i$.  I am having trouble understanding why the variance of $\bar{X}$ is $\sigma^2 / n$.  Here is what I worked out thus far:  $$\begin{gather} \sigma_\bar{X}^2 = E((\bar{X} - \mu)^2) = E(\bar{X}^2 - 2\bar{X}\mu + \mu^2) = E(\bar{X}^2) - 2\mu E(\bar{X}) + E(\mu^2) = E(\bar{X}^2) - \mu. \end{gather}$$  However, I'm not sure what to do with the $E(\bar{X}^2)$.
6225
For random variables $X \in \mathbb{R}^h$, and a positive semi-definite matrix $A$: Is there a simplified expression for the expected value, $\mathop {\mathbb E}[Tr(X^TAX)]$ and variance, $Var[Tr(X^TAX)]$? Please note that $A$ is not a random variable.
81753
From this  on Bayesian Least Mean Square estimation:  $$\mathbf E\left[(\Theta -\hat\theta)^2\right] = \rm{var}(\Theta -\hat\theta) + \left(\mathbf E[\Theta -\hat\theta]\right)^2 = \rm{var}(\Theta) + \left(\mathbf E[\Theta -\hat\theta]\right)^2$$  which is to say that $\rm{var}(\Theta -\hat\theta) = \rm{var}(\Theta)$.  I would like to ask how that comes about?  Thanks.  $\Theta$ is the random variable being estimated, and $\hat\theta$ is a constant estimate.
7965
This should be a fairly straight forward but I still couldn't quite get it. Let $X$ be $n$ by $1$ random vector, $Y$ be another $n$ by $1$ and that $X$ and $Y$ are independent. What is $Var(X^tY)=:V_{XY}$?  I tried generalizing this formula from the univariate case $Var(xy)=E(x)^2Var(y)+E(y)^2Var(x)+Var(x)Var(y)$  except I hit a bit of a roadblock for the 3rd term above. That is, I get $$ V_{XY}=X^tV_YX+Y^tV_XY\\+E[(X^t-\mu_X^t)(Y-\mu_Y)(Y^t-\mu_Y^t)(X-\mu_X)] $$ So... what is it?