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The statement for the independent non-identically distributed random variables do not hold in general. Indeed, it would imply that $X_n/n\to 0$ almost surely and by the Borel-Cantelli lemma we should have $\sum_{n=1}^{+\infty}\mathbb P\{X_n\geqslant n\}<+\infty$. In order to construct a counter-example, we can take $X_n$ taking the values $n$, $-n$ and $... 0 Special about normal distribution is: if$Y$has normal distribution and$Z:=aY+b$(where$a,b$are constants and$a\neq0$) then also$Z$has normal distribution. If we look at non-degenerate cases (as we mostly do) then the standard deviation$\sigma$is positive. If$Xdenotes a random variable then based on it we can define a new one: U:=\frac{X-\mu}... 3 First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we have a S_n/n- b \to a \mathbb{E}(\xi_1) - b a.s.. So a S_n -bn \to - \infty a.s. if and ony if b>a \mathbb{E}(\xi_1) =0, and thus \begin{align*} \exp( a S_n - b n ) \to 0 \text{ iff } b > 0. \end{... 0 I think I can provide a simpler answer: You know that the normal distribution is an even function so you know that the expectation of an odd power must be zero. You know the variance and mean of z^2 and you know that var(z^2)= E[(z^2)^2] - E[z^2]^2 so you know then that E[z^4] = var(z^2)+E[Z^2]^2 = 2+1=3 1 If you know a little bit of calculus, then there is a way to quickly calculate any moment you want for a normal distribution using its Moment Generating Function (MGF). In particular, the MGF for a normal distribution is:M(t)=e^{t\mu+\frac{1}{2}\sigma^2t^2}$$The great property of this function is that E(Z^k)=M^{(k)}(0). So, for your case where \mu=... 0 The derivation of the Normal Distribution (Gaussian Distribution) should answer your question; it also explains why there is a pi and so forth: https://www.sonoma.edu/users/w/wilsonst/papers/Normal/default.html 2 Since the pdf of the standard normal distribution is an even function, it follows that \mathbb{E}[Z^3]=0. And$$ \mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^4e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}x^4e^{-\frac{x^2}{2}}\;dx$$which can be evaluated by setting u=\frac{x^2}{2} and using the properties of the Gamma ... 1 The question, at least in the way you asked it, really doesn't make much sense, because that funny exponential formula is the definition of "normal". Your comments sound like you sort of don't get what I'm getting at here. An analogy: Suppose someone asked this: Q: How do you prove that a triangle has three sides? The answer would be this: A: Huh? Having ... 0 Another way of thinking about the derivative is that it shows the direction of the curve at that point or, more correctly the direction of the tangent line to the curve. If the curve happens to be a straight line, the derivative at any point is just the slope of that line. But where there is a sudden change in direction, such as the 90 degree turn that ... 2 The derivative f'(a) is said to exist at a point a iff the limit \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}} exists. That is, the limit as h\to 0^- (h approaches 0 from the left) should be equal to the limit as h\to 0^+ (h approaches 0 from the right). With this definition, it's easy to see why the derivative does not exist for |x|. I'll ... 0 In the first place your first claim is not really meaningful. Does it mean that if np = n(1-p) = 5 then a normal approximation to the binomial cannot be used? What about 4.9? The point is that you need to specify what you mean by "approximate" before it makes proper sense. Furthermore, this claim is actually false if "approximate" means something like "... -1 This exercise need an additional assumption that ∑ must be a diagnal matrix, so we can ignore all of the elements beside the diagnal. So we can get enter image description here 0 Since both the A and B measurements are of the same thing, why not just add the counts of A and B together to form a new distribution C? This assumes that both A and B's measurement methods are equally reliable, and that more measurements beget a more reliable estimation. If you don't have the counts for each value of X, but you have the number of ... 0 E_x(x^TRx)=\sum_{ij}r_{ij}E_x(x_ix_j)=\sum_{ij}r_{ij}v_{ji}=trace(RV). 0 X = 165+ 6 Z_1 Where Z_n are standard normal random variables. Y = 175 + A Z_1 + B Z_2 a component the depends on X and a component that is independent from X A^2 + B^2 = 64 the variance of Y 6A = cov(X,Y)\\ \rho = \frac {cov(X,Y)}{\sigma_x\sigma_y} = 0.5\\ \frac {6A}{48} = 0.5\\ A = 4\\ B = 4 \sqrt 3\\ Y = 175 + 4 Z_1 + 4 \sqrt 3 Z_2\\ \frac 43 ... 2 Q1: If X and Y are independent normal, and W=aX+bY+c, then W=aX+bY+c is normal, mean aE(X)+bE(Y)+c and variance a^2\text{Var}(X)+b^2\text{Var}(Y). Q2: We use the bilinearity of covariance. We have \text{Cov}(X+Y,X-Y)=\text{Cov}(X,X-Y)+\text{Cov}(Y,X-Y). But \text{Cov}(X,X-Y)=\text{Cov}(X,X)-\text{Cov}(X,Y). This is \text{Var}(X)-\text{Cov}(... 0 1) You are confusing the notation of R programming language with the mathematical meaning. Viz, the theoretical model is$$ \log(y_t) = b_0 + b_1\log(y_{t-1}) + b_2\log(y_{t-12}) + b_3x_1 + e_t, $$and in R notation is written as$$ \log(y_t) \sim \log(y_{t-1}) + log(y_{t-12}) + x_1. $$Hence, 2.312 is simply the OLS estimator of the intercept term b_0. ... 2 It seems that \Phi is meant to be the cumulative distribution function of the standard normal distribution. Since you conclude from \Phi being positive that \Phi^{-1} is positive, I get the impression that you're mistaking \Phi^{-1} to refer to the reciprocal of \Phi. Here \Phi^{-1} denotes the inverse of \Phi. Since \Phi(0)=\frac12, \Phi^{-... 2 To elaborate on @probablyme 's comment, you can use the characteristic function to derive the "common fact". We have that if X \sim N(0,1), then the characteristic function (CF) of X^2 (for any t \in \mathbb{R}) is$$ \begin{align} \varphi_{X^2}(t)&=E \left(e^{itX^2} \right) \\ \\ &= \int_{-\infty}^{\infty} e^{itx^2} \frac{1}{\sqrt{2 \pi}} e^{... 2 SinceX,Y\sim N(0,1)$then$X^2,Y^2\sim \chi^2(1)\implies X^2+Y^2\sim \chi^2(2)$Now use the$\chi^2$distribution's CDF or PDF as required.. 6 Your forgot the Jacobian factor$r$. If you include it, the integral becomes a lot easier to evaluate. 0 If I´m understand it right you want to calculate$P(Y\leq 1175)$, where$Y$is normal distributed as$Y\sim\mathcal N(1200,35^2)$. You can transform the random variable$Y$to$Z=\frac{Y-\mu}{\sigma}=\frac{Y-1200}{35}$. Then$Z$is standard normal distributed:$Z\sim \mathcal N(0,1)$. Thus we have$P(X\leq 1175)=\Phi\left(Z\leq \frac{1175-1200}{35} \right)=...

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Since I am very bad with probabilities, this answer could be totally stupid and, depending on comments, I should delete it. It seems to me that, having $n$ data points $(x_i,P_i)$, the problem is relevant from nonlinear regression; one of the issues is to first obtain reasonable estimates for the parameters $\sigma^2,\alpha,y$. So, take logarithms first and ...

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You should interpret the statement Experience suggests that the standard deviation is more stable than the mean as the assertion that even when the machine is not working properly, the standard deviation of a package weight remains unchanged. Judging from the context, it's a statement about the population $\sigma$, not the sample standard deviation. So the ...

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Your histograms look different because your stats package is binning the data. Normalization of a data set by subtracting its mean and dividing by its standard deviation doesn't really change the shape of the data set; it's only scaling the data and shifting the center to achieve a mean of zero and a sd of 1. If you want to confirm this, do a normal QQ plot ...

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The mean and variance of a sum of normally distributed random variables are, respectively, the sums of the means and variances of those variables. If $X_i\sim\mathcal N(10i, 2^2)$ and all are independently distributed, then : $(\sum_{i=1}^5 X_i)\sim\mathcal N(150, 5\cdot 2^2)$. $$\mathsf P(Z> \tfrac{160-150}{2\surd 5}) \simeq 0.0126{\small 7}$$ TL;DR ...

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Edit: Apparently I could not decipher that the means are $10i$. If I call the subcomponent weights $I_j$, and the total weight $W$, then $$E[W] = E[I_1]+\dotsb+E[I_5] = 10+20+30+40+50 = 150\text{ g}.$$ Assuming the weights are independent, the variance is $$\text{Var}(W) = \text{Var}(I_1)+\dotsb+\text{Var}(I_5) = 5(4) = 20\text{ g}^2.$$ Then \begin{align*}...

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The maximum ordinate of a normal distribution is given by $\frac{1}{\sigma\sqrt{2\pi}}$.

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Note that $x^T \Sigma x = tr(x^T \Sigma x) = tr( x x^T \Sigma)$ by invariance of trace to cyclic permutations and trace of a scalar is itself. Also, note that trace is linear (so we can bring expectations inside the trace). Since $x$ is zero mean, $E[x x^T ] = \Sigma$. Then we have, by the aforementioned, $E[x^T \Sigma x] = E[ tr( x x^T \Sigma) ] = tr(E[x ... 2 Let$\mathbf{X} = (X_1, X_2, \ldots, X_p)$. By definition of$\mathbf{\Sigma}$and matrix multiplication, $$\mathbf{X^T\Sigma X} = \sum_{i=1}^p\sum_{j=1}^p \operatorname{Cov}[X_i, X_j]X_iX_j$$ Since$E[X_i] = 0, i = 1, 2, \ldots, p, the expected value is $$E[\mathbf{X^T\Sigma X}] = \sum_{i=1}^p\sum_{j=1}^p \operatorname{Cov}[X_i, X_j]^2 = \mathbf{1^T(\... 0 r-np \geq z_{1-a}\cdot \sqrt{npq} We can change the inequality to an equality, because we have to square the equation. Squaring is not an equivalent transformation. You can to decide at the end if n has to be \leq or \geq. Taking the square on both sides r^2-2rnp+n^2p^2=z^2\cdot npq p^2n^2-(2rp+z^2\cdot p\cdot q)n+r^2=0 Using the quadratic ... 2 Hint: you have:$$\mathsf P(X\leqslant 0)~=~\tfrac 1 {~3~}~=~\mathsf P(X> 1)$$Or, if Z:=\frac{X-\mu}{\sigma}, so that Z\sim\mathcal N(0,1)$$\mathsf P(Z\leqslant \frac{-\mu}{\sigma})~=~\tfrac 1 {~3~}~=~\mathsf P(Z> \frac{1-\mu}{\sigma})$$3 Details depend on what you are using (tables, software). Using the most standard kind of table, we find that the z such that \Pr(Z\le z)=1-0.015 is given approximately by z=2.17. (Here Z is standard normal.) So we have probability 0.015 in the right tail above z=2.17, and therefore probability 0.015 in the left tail below z=-2.17. Thus ... 1 We have that$$ B_n=A_n+B_n-A_n. $$A_n+B_n\to0 in probability as n\to\infty and A_n\to Z in distribution as n\to\infty, where Z is a standard normal random variable. By Slutsky's theorem, A_n+B_n-A_n converges to 0-Z=-Z in distribution as n\to\infty. Hence, B_n\to Z in distribution as n\to\infty. 0 We have$$1-\text{erf}(x)=\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)/q}$$with q\geq 1. Thus$$1-\text{erf}(x)\geq\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)}$$Now$$P(|Z|>t)=P(Z>t)+P(Z<-t)\\ =2P(Z>t)\\ =2\left(1-\frac{1}{2}(1+\text{erf}(t/\sqrt{2}))\right)\\ =1-\text{erf}(t/\sqrt{2})1-\text{erf}(t/\sqrt{2})\geq \sqrt{\frac{2}{... 1 You are right; you cannot get that end result. It looks like a typo is the sauce of confusion. Taking it slow we have: \begin{align}f_{X+Y}(z)=&~ \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-w^2/2}\frac{1}{\sqrt{2\pi}}e^{-(w-z)^2/2}\operatorname d w \\[1ex] =&~ \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{\pi}}e^{-((... 0 The mean vector is easy: E(X\pm Y) = E(X) \pm E(Y) so \left< (X-Y,X+Y) \right> = (\mu(x)-\mu(y),\mu(x)+\mu(y)) = (<X>-<Y>, <X>+<Y>) $$As to the covariance matrix:$$<(X-Y)^2> - (<X-Y>)^2 = \\<X^2> + <Y^2> - 2<XY> - (<X>)^2 - (<Y>)^2 + 2<X><Y> =\\ [ <X^2> - (&... 0 That is entirely subjective and depends on how much you think each member of each category should get. For example, if you feel each regular player should get three times as much funding as someone who rarely comes, and each "average" player should get twice as much as someone who rarely comes we can come up with the following: Lety_1$be the number of ... 2 Let$X \sim \mathcal N(\mu, \sigma^2)$Then$P(X\leq x+x')-P(X\leq x)=F(x+x')-F(x)=c$In your case$c=0.1$. For a given$x$the equality can be arranged to$F(x+x')=c+F(x)$.$c+F(x)=r$is a real value between$0$and$1$.$F(x+x')=rx+x'=F^{-1}(r)x'=F^{-1}(r)-x$For a given$x$the value of$x'$is$F^{-1}(r)-x$0 In equation (3), Let$E_1$denotes the conditional expectation$E(|X(1)=0)$we use the iterated expectation , that is$E_{1}(X(s)X(t))=E_{1}(E(X(s)X(t)|X(t))\$. In equation (4), since we conditioned to X(t), that means that we know its value , hence X(t) is not random anymore under that assumption: we can remove it from the expectation term as if it was ...

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