For $c>1$, the expected value is the convergent improper integralFor completeness, when $0<c\le 1$ the same positive integral diverges and the expected value is infinite.

Integrating the probability density function from the lower endpoint gives the cumulative distribution functionThe cumulative distribution function is continuous at $x=1$ and tends to one at infinity.

A probability density function must be nonnegative and integrate to one. For $c>0$,At $c=0$ the function is identically zero, while $c<0$ makes it negative on $x>1$. Thus $\boxed{c>0}$ is necessary and sufficient. The resulting law is a Pareto distribution with scale one and shape $c$.

Put $S=X+Y$. Given $X$, its conditional expected value is $3X/2$ and its conditional variance is $X/4$, since $X$ is then fixed. The law of total expectation and law of total variance giveThe two scores are dependent: the law of total covariance gives $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X/2)=q/(2p^2)$. Adding their two variances without that covariance term would give an incorrect total.

The law of total expectation gives $\mathbb{E}Y=\mathbb{E}[X/2]=q/(2p)$. The law of total variance separates the two sources of randomness:Substitution givesThe binomial thinning of a geometric random variable identity supplies a further check: $Y$ itself has a zero-based geometric distribution with success probability $2p/(1+p)$.

Given $X=k$, the score $Y$ has a binomial distribution with success probability $1/2$. Its conditional expected value and variance are $k/2$ and $k/4$. ThusThis is binomial thinning of the first-round count; the second formula uses the identity $\mathbb{E}[Y^2\mid X]=\mathrm{Var}(Y\mid X)+\mathbb{E}[Y\mid X{]}^2$.

Differentiate the probability generating function to obtainThe first derivative is the expected value, and the second gives the factorial moment $\mathbb{E}[X(X-1)]$. HenceIn particular, $\mathbb{E}[X^2]=q(2-p)/p^2$.

Put $q=1-p$. Summing the geometric series in the definition of the probability generating function givesThis is the zero-based geometric distribution, counting failures before the first success.

If the selected coin has bias $p$, the other two biases sum to $3/2-p$. A uniform choice between those two coins therefore has conditional head probability $3/4-p/2$. Average over the posterior distribution, using the previous posterior predictive probability:The observation makes a high-bias coin more likely to have been removed from the remaining pair.

Given the coin, the next toss is independent of its earlier tosses and has that coin's head probability. Averaging using the posterior distribution gives the posterior predictive probabilityObserving two heads raises the probability of having chosen a high-bias coin.

Let $C_j$ identify the selected coin and let $H$ be the event of two heads. The prior coin probabilities are equal, and the conditional probabilities of $H$ are $1/9,1/4,4/9$. By Bayes' theorem,The full posterior distribution over the three biases is $(4,9,16)/29$.

Each position is equally likely to receive any coin, so the law of total probability gives $\mathbb{P}(A_i)=(1/3+1/2+2/3)/3=1/2$. For two specified positions, the ordered pair of distinct coin biases is uniform over six possibilities. Assuming independent tosses conditional on the order,Thus the events are not independent, even pairwise. The covariance of draws without replacement explains the difference: the head indicator random variables have covariance $13/54-1/4=-1/108$. Independent physical tosses given the assigned coins do not imply independence after that hidden assignment is averaged out.

Mutual independence means that for every nonempty $I\subseteq \{1,2,3\}$,Beyond the trivial single-event identities, this requires the three pairwise equalities and the triple equality $\mathbb{P}(A_1\cap A_2\cap A_3)=\mathbb{P}(A_1)\mathbb{P}(A_2)\mathbb{P}(A_3)$. Pairwise independence alone omits that last requirement.

The two definitions of conditional probability giveSubstitution proves Bayes' theorem:The positive probabilities assumed in the question make both conditional expressions defined.

Let $(B_i)$ be a finite or countable partition of a sample space into events of positive probability. The events $A\cap B_i$ are disjoint and their union is $A$. Countable additivity and the definition of conditional probability therefore give the law of total probabilityCells of zero probability contribute zero through $\mathbb{P}(A\cap B_i)$ and can be omitted rather than assigning an undefined conditional probability.

Solving $f_y=0$ gives $y=-1-\alpha x$. Substitution into $f_x=0$ gives $x\{3x+2(1-{\alpha }^2)\}=0$. Therefore $(0,-1)$ is always a critical point. When $\alpha \ne -1$ as well as the printed restriction $\alpha \ne 1$, there is a second critical pointAt $(0,-1)$, the Hessian matrix is $\left(\begin{array}{cc}2& 2\alpha \\ 2\alpha & 2\end{array}\right)$ with determinant $4(1-{\alpha }^2)$. The second-derivative test gives a strict local minimum when $|\alpha |<1$ and a saddle point of a scalar function when $|\alpha |>1$.

The allowed value $\alpha =-1$ requires higher-order terms. With $v=y+1+\alpha x$, completing the square gives the exact identityThe coordinate change is invertible, so the Hessian at a critical point under a change of coordinates preserves the classification. At $\alpha =-1$, this is a cubic degenerate saddle: along $v=0$, the difference from the critical value is $x^3$ and takes both signs. Thus, throughout the printed parameter range,There is no local maximum. When the second point is distinct, it is a local minimum for $|\alpha |>1$ and a saddle point of a scalar function for $|\alpha |<1$: in the $(x,v)$ coordinates its Hessian matrix is $\mathrm{diag}(-2(1-{\alpha }^2),2)$. The printed instruction to split only at $\alpha =1$ misses the negative-parameter cases.

Put $h=x-1$ and $k=y-1$. At $(1,1)$, the value is $5+4\alpha$, the gradient is $(5+4\alpha ,4+2\alpha )$, and the Hessian matrix is $\left(\begin{array}{cc}8& 2\alpha \\ 2\alpha & 2\end{array}\right)$. Thus the quadratic Taylor expansion isThe exact remainder is $h^3$, since the original polynomial has only the cubic term $x^3$ beyond degree two.

The requested partial derivatives areThe last printed derivative has a misplaced superscript on $f$; the second-derivative list and subsequent Taylor expansion use $f_{yy}$.

On a branch with fixed signs of $u,v$ and $uv\ne 0$, use $u^2=x+y$ and $v^2=y-x$. Differentiating these identities, or inverting the Jacobian matrix, givesThe inverse Jacobian matrix is thereforeThese are local partial derivatives. Choosing the signs specifies an inverse function on $y>|x|$; the original map has four preimages there and has no single global inverse. At $u=0$ or $v=0$, the displayed inverse derivatives are not finite.

Differentiating the polynomial coordinate map givesIts Jacobian matrix isThe determinant identifies where the inverse function theorem applies.