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2

We start with the Pearson differential equation: $$f'(x)=-\frac{\left(a_1 x+a_0\right) }{b_2 x^2+b_1 x+b_0}f(x),$$ Define $g(x)=b_2 x^2+b_1 x+b_0$ (using a trick in Diaconis et al.(1991)). Consider (f g)'(x), also written $(f(x) g(x))'=f'(x) g(x) +f(x) g'(x)$. We have $$(f g)'(x)=(-a_0 + b_1 - a_1 x + 2 b_2 x) f(x)$$ We assume that the distribution has ...

2

If you have seen the Moment Generating Function, we can solve this by using it. We have $$\mathbf{E}(e^{tZ}) = e^{\frac{t^2}{2}}.$$ By equating the coefficient of $t^4$, we have $$\frac1{24}\mathbf{E}(Z^4) = \frac18.$$ This gives $\mathbf{E}(Z^4)=3$.

2

$$\mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz$$ set $\frac{z^2}{2}=u$, we have $z\,dz=du$ and $z^3=2\sqrt{2}u\sqrt{u}$, thus $$\mathbb{E}[Z^4]=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}u\sqrt{u}\,e^{-u}du=\frac{4}{\sqrt{\pi}}\Gamma\left(\frac{5}{2}\... 2 Integrate \int_0^\infty x^4 e^{-x^2/2}\; dx by parts using u = x^3, dv = x e^{-x^2/2}\; dx. Or change variables with x = \sqrt{t} and use properties of the Gamma function. 2 Assuming |[X]| means absolute value of floor of X, that is 1 if and only if X \in [-1,0) \cup [1,2). So$$P(X<0||[X]|=1) = \dfrac{P(X \in [-1,0])}{P(X \in [-1,0)) + P(X \in [1,2))}$$Also useful will be symmetry, so P(X \in [-1,0)) = P(X \in [0,1)). 2 If X\sim \chi^2_n then \Pr(X\ge 0) =1. But \Pr(X_1^2+X_2^2 - X_3^2 \ge 0 ) <1, because the probability that X_1^2+X_2^2 is small and X_3^2 is large is not 0. 1 The question asks you to verify that E[\bar{X}^2]\neq \mu^2. The expansion E[\bar{X}^2] gives:$$ E\Big[\bar{X}^2\Big]=\frac{1}{n^2}E\Big[\Big(\sum_iX_i\Big)^2\Big]=\frac{1}{n^2}E\Big[\sum_iX_i^2+2\sum_{i<j}X_iX_j\Big]. $$Independence only gives E[X_iX_j]=E(X_i)E(X_j)=\mu^2 for i<j. Independence doesn't make expectations of the cross terms go ... 1$$ E\bar{X}_n^2 = var(\bar{X}_n)+E^2\bar{X}_n = 1/n+\mu^2 >\mu^2. $$So the simplest unbiased estimator would be$$ \bar{X}^2_n-1/n $$1 M and x are independent? \int_0^1Mx=M\int_0^1x=M/2 so I\sim N(0,1/4). 1 I doubt there can be a definite answer unless we find the person who wrote it there and asked him or her. I am guessing that G refers to Gaussian. Could \mu+C(\sigma) be a confidence interval? and the person might want to know what is the image of this interval under exponential map? 1 The variable X is not independent of variable X. Thus the correlation coefficient is unequal to zero. The variance of the sum of two dependent, normally distributed random variables (X,Y) is \sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2+2\rho \sigma_X\cdot \sigma_Y Setting Y equal to X. \sigma_{X+X}^2=\sigma_X^2+\sigma_X^2+2\rho_X \cdot \sigma_X\cdot \... 1 If X \sim N(\mu_X ,\sigma_X) and Y \sim N(\mu_Y ,\sigma_Y) then$$X + Y \sim N(\mu_X + \mu_Y,\; \sigma_X^2 + \sigma_Y^2 + 2\sigma_{X,Y})$$We have$$X + X \sim N(\mu_X + \mu_X,\; \sigma_X^2 + \sigma_X^2 + 2\sigma_X^2)$$Indeed$$2X \sim N(2\mu_X ,\; 4\sigma_X^2 )$$Note$$2\sigma_{X,Y}=2\operatorname{Cov}(X,Y) 2\sigma_{X,X}=2\operatorname{Cov}(X,...

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The two word answer is "polar coordinates". In more detail, let $f:S^{n-1}\to\Bbb R$ be a continuous function. Then \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ... 1 Yes: the normal distribution cares only about the distance between the mean and the point. Note that in g(x \mid x_p), the mean is x_p now, not x, so the distance between the mean in the two cases are the same and the variance in the two cases are the same. (In this symmetric situation, the algorithm is commonly called just Metropolis, rather than ... 1 I consider the two variable case and take into account that the means of X_1 and X_2 are 0. The pdf of the univariate untruncated distribution of X_1 is f_{X_1}(x_1)=\frac{1}{\sqrt{2\cdot \pi \sigma_1^2}}\cdot e^{-\frac12 \left[\frac{x_1^2}{\sigma_1 ^2}\right]}  The cdf of the univariate untruncated distribution of X_1 is F_{X_1}(x_1)=\int_{-\... 1 The pdf of Y is obtained by taking the joint pdf of (X,Y) and marginalizing X out. That is:f_Y(y)=\int_{-\infty}^\infty f_{X,Y}(x,y) dx.$$The joint pdf of (X,Y) is the product of the conditional pdf f_{Y|X}(y|x) and the pdf of X, f_X. (If this seems weird to you, it is basically analogous to the familiar identity P(A \cap B)=P(A \mid B) ... 1 Let X_1,..,X_n i.i.d \mathcal{N}(\mu, \sigma^2) each one,hence by definition$$ \sum_{i=1}^n\left(\frac{X_i - \mu}{\sigma} \right)^2 \sim \chi^2_n \, . $$Now, in order to estimate the variance of the Normal distribution, you are using (variation of)$$ S^2 = \frac{1}{n}\sum_{i=1}^n(X_i - \bar{X})^2, $$thus$$ \sigma^2 S^2 = \frac{\sigma^2}{n}\sum_{i=...

1

The new density is again truncated normal at $t$ with new parameters $\frac{\sigma^2\mu_0+\sigma_0^2 x}{\sigma^2+\sigma_0^2},\sigma^2\sigma_0^2/(\sigma^2+\sigma_0^2)$. The new normalising constant is $$\Phi\left( \frac{t-\frac{\sigma^2\mu_0+\sigma_0^2 x}{\sigma^2+\sigma_0^2}}{\sigma^2\sigma_0^2/(\sigma^2+\sigma_0^2)}\right )$$ So the interesting thing ...

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