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6

Your forgot the Jacobian factor $r$. If you include it, the integral becomes a lot easier to evaluate.


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}(...


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^{-...


1

In $\Pr(X-2Y<0)=\Pr\left(Z \leq \frac{0-0}{5}\right)$ you need $\sqrt 5$ rather than just $5$ in the denominator. You divide by the standard deviation, not by the variance. For the first question you need $W\sim\mathcal{N}(-2,10)$. For the normal distribution you specify the expected value and the variance to say which normal distribution it is. ...


1

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 ...


1

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 ...


1

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^{...


1

Since $X,Y\sim N(0,1)$ 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..



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