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4

As the covariance matrix is not inversible, there is a relation (assuming $X_1\neq 0$ and using the fact that the expectation of both random variables is $0$) $$ X_1 = aX_2 $$ and hence $$ E[X_1^2 X_2^2] = a^2E[X_1^4] = 3a^2\sigma_1^4 $$ where $\sigma_1^2$ is the variance of $X_1$, that is $Q(1,1)$. To get rid $a^2$ write the same thing with $X_2$: $$ ...


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. $


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 ...


3

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


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

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 $1.0$ ...


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} ...


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 ...


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 ...


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 $mpg_2$: $P$($mpg_1$ $\le$ $MPG$ $\le$ $mpg_2$) $= 0.60$.


2

Calculation by using the z-score The equation is $P(X > x)=0.3$ We know, that $P(X > x)=1-P(X \leq x)$ Thus the equation with the standard normal distribution is: $1-\Phi \left( \frac{x-\mu}{\sigma}\right)=0.3$, with $Z=\frac{X-\mu}{\sigma}$ Z is standard normal distributed: Z $\sim \mathcal N(0,1)$ $1-\Phi \left( \frac{x-30}{8}\right)=0.3 \quad ...


1

You are taking $X=\sigma_xZ+\mu_x$ and $Y=\sigma_yZ+\mu_y$. They are not independent. You should take $X=\sigma_xZ_1+\mu_x$ and $Y=\sigma_yZ_2+\mu_y$ where $Z_1$ and $Z_2$ are iid with standard normal distribution.


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

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?


1

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

$\checkmark$ 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) = ...


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

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?


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 ...


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.


1

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 ...


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 ...


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

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 $$\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{.383}{.6826}=0.561$$


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

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 $Z = Y_1 - Y_2 \sim N(-4, 10)$.


1

Consider $$\frac{CDF_1(x)}{CDF_2(x)} = P.$$ We have $$CDF(x) = \frac{1}{2} \left[1 + erf\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]$$ and $$\frac{d}{dx}CDF(x) = \frac{1}{\sigma 2\sqrt{2}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}.$$ Using the above two equations, write $$CDF_1(x) = P \times CDF_2(x)$$. Differentiate on both sides with respect to $x$ and solve ...


1

First: $\frac{1}{\sqrt{n}}(T - n\mu) \sim N(0, \sigma^2)$ does not imply $T\overset{H_0}{\sim} N(\mu_0, \sigma^2)$ since $\mu_0$ pertains to the mean of a single $X_i$ not the sum. Instead $\frac{T}{n}\sim N(\mu_0, \frac{\sigma^2}{n})$ under the null hypothesis. Also, since your data are already normally distributed, you don't need to invoke the CLT or any ...


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

Notations: Let $F$ and $f$ be the cdf and pdf of $X$ respectively. Let $\Phi$ and $\phi$ be the cdf and pdf of a standard normal variable. Part 1: Justification First of all can you calculate $P(a<X<b|X>72)$? Yes you can! It is $\displaystyle \frac{P(a<X<b \cap X>72)}{P(X>72)}$. Now you can treat $Y:=X|X>72$ as a random variable ...



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