The centered design gives $\hat{\beta }=\beta +S_{xx}^{-1}{\sum }_i{x}_i{ϵ}_i$. By independence of the errors, their off-diagonal covariances vanish. Thus variance additivity for independent random variables givesA larger spread of predictor values decreases the slope variance for fixed error variance.

Differentiate the ordinary least squares objective in a. Its normal equations for linear least squares areUsing ${\sum }_i{x}_i=0$ and $S_{xx}={\sum }_i{x}_i^2>0$ givesThe Hessian matrix of the objective in $(a,b)$ is $2\mathrm{diag}(n,S_{xx})$, which is positive-definite, so this is its unique global minimum. For the observed estimates, replace $Y_i$ by $y_i$.

For a geometric distribution on positive integers, the probability generating function is $ps/[1-(1-p)s]$. Conditional averaging over the red die yieldsAs a power series, its exact domain of absolute convergence isIndeed the slowest decaying component has failure probability $5/6$, so $\mathbb{P}(T=t)\ge (1/36)(5/6{)}^{t-1}$. At $|s|=6/5$, the terms of the power series fail to tend to zero; outside that disk they grow. All six component series converge inside it. In particular, the probability generating function is defined on the usual real interval $0\le s\le 1$, with $G_T(1)=1$. The rational expression beyond the convergence disk is only an algebraic continuation, not the defining expected value.

By the law of total probability,Bayes' theorem therefore gives the conditional probability mass functionThe six probabilities sum to one. A small red outcome is more likely after an immediate success because more white faces can meet its threshold.

Let $R$ denote the red die. Conditional on $R=r$, each white roll succeeds independently with probability $p_r=(7-r)/6$. Consequently $T\mid R=r$ has the geometric distribution on $\{1,2,\dots \}$ and expected value $1/p_r=6/(7-r)$. The law of total expectation givesThis is $2.45$ rolls; the averaging must be over conditional expected values, not over success probabilities before taking a reciprocal.

On the product support $[0,1{]}^2$, the joint probability density function factors asThe equality also holds outside the support with both sides zero. The probability density function factorization criterion proves that $U$ and $V$ are independent random variables.

Obtain each marginal density by integrating the joint probability density function over the other variable:Each marginal density is zero off its displayed interval and has total integral one.

A joint probability density function integrates to one. HeresoThis value also makes the probability density function nonnegative everywhere.

Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The standard normal distribution is symmetric and assigns no probability mass to zero. Thus $\mathbb{P}(B)=1/2$, and its equal two-sided tails giveThereforeso the two events are independent. The permitted approximation gives $\mathbb{P}(A)\approx 0.05$ and $\mathbb{P}(A\cap B)\approx 0.025$, but independence is exact.

Absolute integrability now requiresFor $\alpha \le 0$, the integrand is bounded by $z{e}^{-z^2/2}$, whose integral is finite. For $\alpha >0$, once $z^2\ge 1/\alpha$, the exponent is at least $\alpha z^4/2$, so the integral diverges. HenceAgain the zero follows from the odd function and symmetry; outside the integrable range both signed parts diverge, so the expected value is undefined.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

The homogeneous characteristic equation of a linear recurrence isThe repeated characteristic root of a linear recurrence $-1$ supplies the independent terms $(-1{)}^n$ and $n(-1{)}^n$. A constant particular solution is $1/2$, since $1/2=-2(1/2)-1/2+2$. Thus $u_n=1/2+(A+Bn)(-1{)}^n$.

The scan prints $u_2=0$, not $u_1=0$. These initial conditions are consistent. They give $A=1/2$ from $u_0=1$, then $0=1/2+A+2B$, so $B=-1/2$. ThereforeIn particular $u_1=1/2$ and $u_2=0$. Direct substitution verifies the recurrence relation. This is an example of nonconsecutive initial values for a linear recurrence; replacing the printed subscript by 1 would change the answer.

For the homogeneous linear recurrence relation, try $u_n=r^n$. Its characteristic equation of a linear recurrence isThe distinct roots of a polynomial $1,-4$ give the complementary terms $A+B(-4{)}^n$. A constant cannot absorb the forcing because $1$ is already a root of a polynomial of the characteristic equation; try a particular solution $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The general solution is $u_n=A+B(-4{)}^n+n/5$.

The initial conditions give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and

Summing over the last digit yields the marginal probability mass functionTogether with the other marginal probability mass function, this givesThe factorization holds on the whole support and with zeros elsewhere, proving that $X$ and $Y$ are independent. The conditional distribution in ii likewise does not depend on the permissible value $x$.

Summing the joint probability mass function over $y$ with the geometric series gives the marginal probability mass functionDivide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:It is the geometric distribution counting failures before the first success, with success probability $1-q^{10}$. For the other nonnegative integers $x\ge 10$, $\mathbb{P}(X=x)=0$; the elementary conditional probability given $X=x$ is undefined and supplies no additional case.

The printed numerical example reverses the digits and contradicts both definitions. With the stated last-digit definition, $[Z]=17$ gives $X=7$, $Y=1$. Use those definitions throughout.

Put $q=e^{-\lambda }\in (0,1)$ and $N=\lfloor Z\rfloor$. The exponential distribution givesThe Euclidean division $N=10Y+X$ has unique remainder $X\in \{0,\dots ,9\}$ and quotient $Y\in \{0,1,\dots \}$. Thus the joint probability mass function isand zero otherwise. This is decimal digit decomposition of a geometric random variable. Following the erroneous example instead would require different definitions and would not answer the stated question.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

For an outcome $\omega =(i,j)$, the two random variables are the coordinate mapsTheir marginal probability mass functions are $p_X(i)=p_Y(j)=1/6$ on $\{1,\dots ,6\}$, and their joint probability mass function satisfiesOff that support, the same factorization holds with zeros. Hence $X$ and $Y$ are independent random variables, by the joint probability mass function criterion for independence.

Use the ordered outcomes of the two rolls as the sample space:Thus each elementary outcome has probability $1/36$. This is the usual probability space for two independent fair dice.