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The Statistics Toolbox has a function mvnrnd(see its documentation) to generate jointly Gaussian random variables with specified means and covariance matrix: N = 10; % desired number of samples of each variable mu = [10; 10]; % vector of means cov = [3 1; 1 3]; % covariance matrix samples = mvnrnd(mu, cov, N); If you want to do it manually, you can ...


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$\displaystyle \sum_{i=1}^N \left(\frac{x_i-\mu}{\sigma}\right)^2$ has a $\chi_N^2$ distribution (chi-squared with $N$ degrees of freedom) as as the sum of $N$ independent standard normal random variables. So if $\mu=0$ and $\displaystyle s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2$ then $N s_2^2 / \sigma^2$ also has a $\chi_N^2$ distribution.


1

You did it all right, except for the final fraction. The joint event is: $$A \cap B \equiv (1 \le X \le 3) \cap (0 \le X \le 4)\equiv (1 \le X \le 3)\equiv A$$ (Notice that this holds because here $A \subset B$) Hence $$\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{.383}{.6826}=0.561$$


2

You need to conduct a hypothis test for the variance not for the mean. The confidence interval $9\pm 1.5$ is a confidence interval for the mean and will not help you to draw conclusions about the variance! Accordingly, you do not need the $t$-distribution but the $\chi^2$-distribution. Formally In order to conclude whether $σ^2$ has been reduced below $1.0$ ...


3

Fact 1: If $X\sim N(0,a)$ then $\mathbb{P}(|X|>\epsilon)\leq \frac{1}{\epsilon}\sqrt{\frac{2a}{\pi}}\exp[-\epsilon/2a]\leq \exp[-\epsilon/2a]$, whenever $a\leq \epsilon^{2}\pi/2$. Fact 2: If $t\to 0$ then $t^{-b}e^{-c/t}\to 0$ for any fixed $b,c>0$. In particular $e^{-c/t}<t^{b}$ for sufficiently small $t$. We know that $a_{n}\to 0$. So there ...


2

Suppose that $$\hat{f}(\xi) = \sqrt{\hat{g}(\xi)} \cdot h(\xi)$$ for some function $h: \mathbb{R} \to \{-1,1\}$. Since $\hat{f}$ is itself a characteristic function, we know that $\tilde{f}(0)=1>0$. Hence, $h(0)=1$. On the other hand, $$h(\xi) = \frac{\hat{f}(\xi)}{\sqrt{\hat{g(\xi)}}}$$ is a continuous function as $\hat{f}$, $\hat{g}$ are continuous ...


1

Yes, one knows from general principles that, for each fixed $t$, the random couple $(Y_t,Z)$ is a linear functional of the gaussian process $(Y_s)_{0\leqslant s\leqslant1}$ hence $(Y_t,Z)$ is normal. Let $C$ denote the covariance of the process $(Y_s)_{0\leqslant s\leqslant1}$, that is, $C(u,v)=E(Y_uY_v)$ for every $(u,v)$, then $(Y_t,Z)$ is centered, ...


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This is the same as $P \left(X_1-X_2>0\right)=\int_0^{\infty } \left(f_1\left( x\right)-f_2\left( x\right)\right) \, dx$ $$ =\frac{1}{2} \left(\frac{\text{erf} \text{$\mu $2}}{\sqrt{2} \sigma }-\frac{\text{erf} \text{$\mu $1}}{\sqrt{2} \sigma }\right)$$


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Find the $z$ value of a normal distribution with stdev=100 at $0.2$. $z_{0.2}*100kg + \mu = 22$ ... that gives you $\mu$. $z_{x}*100kg + \mu = 20$ ... find $x$ to solve.


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Hints: For a standard normal distribution with mean $0$ and standard deviation $1$, at what value is $20\%$ of the distribution above that value. How many standard deviations above the mean does that represent? If the standard deviation were $0.1$ then how far below $22$ would the mean be? What is the probability of at least $20$ given that calculated ...


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The distribution of the sum of the two independent normally distributed variables $X \sim N(\mu_X, \sigma_X^2)$, $Y\sim N(\mu_Y, \sigma_Y^2)$ is normally distributed, more specifically if $Z = X + Y$ then $Z \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$, see e.g. the wiki for a proof. In your case that would result in $Z = Y_1 - Y_2 \sim N(-4, 10)$.


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Basically, the accumulation of a large number of very small independent contributions will be approximately normal (according to the Central Limit Theorem). For example, take heights of people: if you're talking about a homogeneous population, all adults of the same gender, so that the height differences are the result only of small environmental and ...


2

Consider: Experiment: Randomly select one compact car Random Variable $MPG$: $M$iles $P$er $G$allon Possible Values $mpg$: [$0$ mpg, $600$ mpg] Determine $mpg_1$ and $mpg_2$: $P$($mpg_1$ $\le$ $MPG$ $\le$ $mpg_2$) $= 0.60$.


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The mean +/- 1σ is about 68% of the total. So you want less than +/- 1σ. Look here: http://upload.wikimedia.org/wikipedia/commons/7/75/Standard_score_and_prediction_interval.png If you read the chart to get 60%, you'll see you need about +/- 0.84σ.


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Consider: Experiment: Randomly select one American woman Random Variable $CL$: $C$holesterol $L$evel Possible Values $CL$: [$0$ mg/dl, $400$ mg/dl] Determine $cl$: $P$($CL \ge$ $cl$) $=$ $0.15$


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Your handling of 1) is correct. It works fine because function $M$ is strictly increasing. $X=M(T)$ is completely determined by $T$ and $M(t)\geq 0.9$ is true for every value $t\geq12$. This allready justifies the conclusion: $$P(X=M(T)\geq0.9\mid T\geq12)=1$$ Working alternatively with conditional probability would lead to $$P(X\geq0.9\mid ...


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As hinted in the previous post and in the comments the amount of money you have after $10$ years can be written as $$M = 2^{\sum_{i=1}^{10} X_i}$$ where $X_i$ are independent random variables which takes the values $x_i =\{-1,0,1,2\}$ with probabillity $p_i=\{2/8,3/8,2/8,1/8\}$. The expected value of each $X_i$ is given by $$\mu = \sum_{i=1}^4 p_i x_i = ...


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First generate a vector $\mathbf{x}$ with independent $N(0,1)$ normal random components: $$x_i \sim N(0,1).$$ Find the Choleski decomposition of the covariance matrix $\Sigma$: $$\Sigma = L L^T .$$ Then $\mathbf{y} = L^T\mathbf{x}$ is jointly normal with covariance matrix $$E(\mathbf{y}^T \mathbf{y})=E(\mathbf{x}^TLL^T \mathbf{x})=\Sigma E(\mathbf{x}^T ...


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Hint: If $X\sim N(\mu, \sigma^2)$ then $Z:= \frac 1\sigma (X-\mu)$ is also a normal variable. What are its mean and variance?


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No normal (even continuous) distribution would fit your evolution rules, which are clearly discrete (if only because $P(X_{t+1} = X_t) = 0 \not= 3/8$. With your evolution probabilities, the best you can do IMO is to enumerate all possible values of the stock in year $n$ for $n \le 10$ and then compute the probabilities each year using a probability tree ...


1

Based on the following observations $Y$ takes values in $[0,1]$, $Φ(\cdot)$ is strictly monotone increasing and therefore invertible. Let $Φ^{-1}$ denote it's inverse, the CDF $F_Y(y)$ of $Y$ is given by $$F_Y(y)=P(Y\le y)=P(Φ (X)\le y)=P(X\le Φ^{-1}(y))=Φ(Φ^{-1}(y))=y$$ for $y\in[0,1]$. Thus $$f_Y(y)=\frac{\partial }{\partial y}F_Y(y)=(y)'=1$$ for $y ...


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Your inversion formula is wrong. From $$ u = \frac{1}{1+\exp(-\frac{x_i-\bar{x}}{\sigma})}, $$ you can solve \begin{align} u_i &= \frac{1}{1+\exp(-\frac{x_i-\bar{x}}{\sigma})}\\ u_i(1+\exp(-\frac{x_i-\bar{x}}{\sigma})) &=1\\ u_i+ u_i \exp(-\frac{x_i-\bar{x}}{\sigma}) &=1\\ u_i \exp(-\frac{x_i-\bar{x}}{\sigma}) &=1 - u_i\\ ...


1

Here is how you would plot a normal distribution in Matlab for different $i = \mu$ and $j = \sigma$. x = linspace(-10, 10, 5000); for i = 0:1:5 for j = 1:1:6 n = 1/(j^2*sqrt(2*pi))*exp(-(x - i)^2/(2*j^2)); hold on plot(x,n) end end So I plotted $\mu = 0, 1, \ldots, 5$ against $\sigma = 1,2,\ldots, 6$. If you want to plot ...


1

Let $g_t(x)$ denote the PDF of $X_t$ at point $x$, then, by homogeneity, $\|X_t-X_s\|_{TV}=4d\left(\frac{s}t\right)$ where, for every $s$ in $(0,1)$, $$d(s)=\int_{x(s)}^\infty (g_1(x)-g_s(x))\,\mathrm dx,$$ and $x(s)$ is the point where the two densities coincide, that is, $x(s)$ is your $x(s,1)$. Equivalently, once again by homogeneity, ...


3

Consider: Experiment: Randomly select one power supply Random Variable $L$: Power Supply $L$ifetime Possible Values $l$: [0 hours,1300 hours] Determine: $P$($7,225 \le$ $L$ $\le$ $7,550$) $=$ $P$($\dfrac{7,225-6,500}{750}$ $\le$ $Z$ $\le$$\dfrac{7,500-6,500}{750}$) $=$ $P$(0.97 $\le$ $Z$ $\le$ $1.33$) $= 0.9082 - 0.8340 =0.0742. $


0

Let $F$ denote the CDF of the uncensored weight, then the median $m_x$ of the weight censored below the value $x$ solves $F(m_x)=\frac12(1+F(x))$.


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OP asks: Prove that $Y = \frac{X_1+X_2*X_3}{\sqrt{1+X_1^2}}$ obeys normal distribution How about a proof that it doesn't? Here is a quick Monte Carlo check of the empirical pdf of $Y$: which is decidedly not Normal. What is the source of your question?


0

X is uniformly distributed, $X\in \langle 0,10\rangle \Rightarrow E(x) = \frac{0+10}{2}=5,\,D(x)=\frac{(10-0)^2}{12}=\frac{25}{3}$ Average values ​​at 100 experiments: $\mu_{100}=E\left(\frac{x_1+x_2+\cdots+x_{100}}{100}\right)=E(x)=5$ $D_{100}=D\left(\frac{x_1+x_2+\cdots+x_{100}}{100}\right)=\frac{D(x)}{100}\doteq 0.08333\Rightarrow \sigma \doteq 0.2887$ ...


0

If $ X \choose y$ is normally distributed, then $P(X|Y=y)=\mu _x+\rho \frac{\sigma _x}{\sigma _y}(Y-\mu _y) \quad (1)$ Or other way round. If $ X \choose y$ is normally distributed, then $P(Y|X=x)=\mu _y+\rho \frac{\sigma _y}{\sigma _x}(X-\mu _x) \quad (2)$ You can substitue: $\mu _x=a,\mu _y=b,\frac{\sigma _x}{\sigma _y}=c$. After you have transformed ...


1

You may want to look at the Berry-Esséen theorem (and variants, for other metrics: the Berry-Esséen theorem is essentially a quantitative version of the Central Limit Theorem, phrased in terms of Kolmogorov distance between $Z_n$ and the corresponding normal random variable. There are extensions and generalizations, e.g. to multivariate random variables — ...


1

$\checkmark$ Yes, that looks long winded but correct. The support for $Y$ must be $(-\infty, 0)\cup(0,\infty)$.   Or equivalently: $(-\infty,\infty)\setminus\{0\}$ The transformation is: $$\begin{align} f_Y(y) & = f_X\circ g^{-1}(y)\cdot \bigg|\frac{\mathrm d g^{-1}(y)}{\mathrm d y}\bigg| & : f_X(x) = ...


0

I think I'm fine with it now. We calculate: $ \int\limits_{0}^{\infty} e^{\frac{x}{2}(2it - 1)} \frac{1}{\sqrt{x}} dx $ by substituting $ s^2 = x(-2it + 1), ds = \sqrt{1-2it}\frac{1}{2\sqrt{x}}dx $ and we get that it equals $ \int\limits_{0}^{\infty} e^{-s^2/2} \frac{2}{\sqrt{1 - 2it}}ds = 2\sqrt{2\pi}/\sqrt{1-2it} $ And the characteristic function: $ ...


0

A normally distributed r.v. has pdf $\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$, and therefore the characteristic function of $X^2$ where $X\sim\mathcal{N}(0,1)$ is $$ \mathbb{E} e^{itX^2} = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{it x^2}e^{-\frac{x^2}{2}} dx = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{(it-\frac12) x^2} dx= ...


2

You messed things up when you used the density of $X^2$ and wrote $\mathrm e^{\mathrm itx^2}$ in the integral. Either use the density of $X$ and write $\mathrm e^{\mathrm itx^2}$ in the integral, or use the density of $Y=X^2$ and write $\mathrm e^{\mathrm ity}$ in the integral. Personally, I find the former option more systematic and less error-prone than ...


2

By independence, $P_{(X,Y)}=P_X\otimes P_Y$, and hence $$ \phi_{XY}(t)=\int_{\mathbb{R}^2}\mathrm{e}^{itxy}\,P_{(X,Y)}(\mathrm dx,\mathrm dy)=\int_{\mathbb{R}^2}\mathrm{e}^{itxy}P_X\otimes P_Y(\mathrm dx,\mathrm dy). $$ The complex version of Fubini's theorem now allows us to write this as an iterated integral $$ \begin{align} ...


0

Revised per OP Comment If you only have 10 measurements and an assumed standard error, then you can use a Bonferroni Correction to develop simultaneous confidence intervals with a guaranteed maximum family-wise error rate (assuming that your errors are well approximated by a normal distribution): Select your desired overall Type I error rate $\alpha$ We ...


0

If you map a vector space into a vector space of higher dimension, the image is contained in proper subspace of dimension at most the dimension of the original vector space. Thus mapping into a higher dimension is equivalent to mapping into a vector space of equal or lower dimension.


0

If $X_i$ is the height of the $i^{th}$ sampled male and $Y_j$ is the height of the $j^{th}$ sampled female, then - assuming these random variables are normally distributed - you know that $$ X_i \sim \mathcal{N}(180,4), ~~~ Y_j \sim \mathcal{N}(170,5) $$ where I'm writing $\mathcal{N}(\mu,\sigma)$ to denote a normal random variable and its parameters. One ...


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The thing to know is that if $$X_1, X_2, \ldots, X_n \overset{\mathrm{iid}}{\sim} \operatorname{Normal}(\mu, \sigma^2)$$ is a sample of independent and identically distributed normal random variables with mean $\mu$ and variance $\sigma^2$, then the sample mean $$\bar X = \frac{X_1 + X_2 + \cdots + X_n}{n} \sim \operatorname{Normal}\Bigl( \mu, ...


0

Let $\overline{X}$ be the random variable you get by taking a random sample of $16$ males and finding their average height. Then $X$ is normally distributed, with mean $\mu_X=180$, and variance $\frac{\sigma_X^2}{16}=1$. Now all you need to do is to compute the probability that a normal with mean $180$ and variance $1$ is less than $178$. For the ...


0

The average of the two heights is $(X+Y)/2$. Its expected value is $\operatorname{E}\left( \frac{X+Y}2 \right)= \frac{180+170}2$. Its variance is $$ \operatorname{var}\left(\frac{X+Y}2\right) = \frac 1 4 \operatorname{var}(X+Y) = \frac 1 4 \left(\operatorname{var}(X)+ \operatorname{var}(Y) \right) = \frac 1 4 (16+25). $$ The average of the two is normally ...


1

The male is more than $5$ cm taller than the female if $X\gt Y+5$. I would then probably let $W=X-Y$. We want $\Pr(W\gt 5)$. From a standard result, we have that $W$ has normal distribution, mean $\mu_X-\mu_Y$, and variance $\sigma_X^2+\sigma_Y^2$. Now that we know the mean and variance of $W$, we can find $\Pr(W\gt 5)$ in the usual way. Second ...


1

In general, one would expect that a smaller sample would result in a larger (say $95\%$) confidence interval. However, in the situation you describe, one could easily end up with a smaller (that is, better) confidence interval. That is because you were using the $t$ distribution. Discarding the outliers may have decreased the sample variance enough to ...


1

Ad a) The random variable is binomial distributed, $X \sim B(400,0.4)$. Ad b) Now you can approximate the binomial distribution by the normal distribution. You take the mean and the standard deviation of the binomial distributed random variable. $\mu=0.4\cdot 170$ and $ \sigma=\sqrt{ \sigma ^2 } =\sqrt{n\cdot p \cdot (1-p)}=\sqrt{400\cdot 0.4\cdot 0.6}$ ...


1

You want $\Pr(Z\lt 0.35)-\Pr(Z\le -1.1)$. The table does not have direct information about $\Pr(Z\le a)$ for negative $a$. But by symmetry we have $\Pr(Z\le -1.1)=\Pr(Z\ge 1.1)$. Note that $\Pr(Z\ge 1.1)=1-\Pr(Z\lt 1.1)$. Now you have all the needed components.


2

Use the standard result: Joint distribution of two random variable is Gaussian implies that the marginal distributions of the random variables are also Gaussian. In the above problem, let $\mu =(\mu_1,\mu_2)$ and $\Sigma_x=\begin{bmatrix}\Sigma_1 & \Sigma_{12} \\ \Sigma_{12} & \Sigma_2 \end{bmatrix}$. Then the distribution of $(X_1+X_2)$ is ...


1

Romberg integration is a very relevant method and the code given in Numerical Recipes is pretty good, so I suggest you go ahead. Personally, I should however start with a change of variable $s=\frac 1x$ which transform the integral $$\int\frac{e^{-\frac{z^2}{8 s^2}-\frac{s^2}{2}}}{\sqrt{2 \pi }}ds=-\int\frac{e^{-\frac{x^4 z^2+4}{8 x^2}}}{\sqrt{2 \pi } ...


1

The distribution of the variable is $$ X = 1_{U<p} X_1 + 1_{U\ge p} X_2 \sim 1_{U<p} N(0, \hat\Sigma_1) + 1_{U\ge p} N(0, \hat\Sigma_2) $$ (with $U\sim U[0,1]$ is independent of the rest) which as expected value $0$ and second moment \begin{align} E[X_iX_j] &= E[(1_{U<p} X_{i,1} + 1_{U\ge p} X_{i,2})(1_{U<p} X_{j,1} + 1_{U\ge p} X_{j,2})] ...


1

I don't think there's a purely analytical way to do this, but perhaps I can offer something a little more systematic than guess and check: Lets first consider the much simpler problem of a mean $0$ normal RV with variance $9$. In that case, we know from lookup tables that $b\approx 1.96\times 3$ will solve our problem. However, the shift upwards by two ...



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