In the probability density function convention used here, a random variable $X$ has a nonnegative probability density function $f$, with ${\int }_{\mathbb{R}}f(x)\mathrm{d}x=1$, such that for every Borel set $A$,In particular, its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)\mathrm{d}t$ and individual points have probability zero. More precisely, this property is the statement that the distribution is an absolutely continuous probability distribution; continuity of the cumulative distribution function alone is a weaker property and need not imply a probability density function.

Let $T_2$ be the time of the second occurrence of $3$, whether or not it is adjacent to the first. On every digit sequence, completion of a consecutive pair requires at least two occurrences of $3$, so $Y⩾T_2$.

The waiting time for the first $3$ has mean $3$. Once that digit has appeared, the fresh independent digits give another waiting time with the same mean for the next $3$. Thus $\mathbb{E}[T_2]=3+3=6$, the mean of the corresponding negative binomial distribution. Monotonicity of expectation now proves $\mathbb{E}[Y]⩾6$, without computing the waiting time for the consecutive pair itself. The argument remains valid even before establishing that $Y$ has finite mean, since these waiting times are nonnegative.

There is no forbidden pair in a sequence of length zero or one, so $p_0=p_1=1$. For $n⩾2$, divide acceptable sequences into two disjoint cases. If the first digit differs from $3$, its probability is $2/3$, and the remaining $n-1$ independent digits must avoid the pair. If the first digit is $3$, the next must differ from $3$, which has combined probability $2/9$, and the remaining $n-2$ digits must avoid the pair. The boundary between this prefix and the remaining digits cannot create a forbidden pair, because the prefix ends in a digit other than $3$. ConsequentlyThe characteristic polynomial of a linear recurrence is $r^2-\frac{2}{3}r-\frac{2}{9}$, with distinct rootsWriting $p_n=A{r}_{+}^n+B{r}_{-}^n$, the initial conditions give $A+B=1$ and $A{r}_{+}+B{r}_{-}=1$. ThereforeFor example, $p_2=8/9$, since only one of the nine equally likely pairs is forbidden. This is the case with success probability $1/3$ of the avoidance recurrence for two consecutive successes.

For $X=m$, the first $m-1$ digits must all differ from $3$ and the next digit must be $3$. Independence givesThus $X$ has the geometric distribution counting trials up to and including the first success, with success probability $1/3$. For completeness, its mean follows from the tail-sum formula:The sum of the geometric series also proves finiteness of this expectation.

There is no such value of $p$ in the specified interval. Indeed, $\mathbb{P}(Y=1)=p{e}^{-p}>0$, and the joint probability mass function givesThis contradicts independence for every $0<p<0.8$. Equivalently, $(1,2)$ has joint probability zero but the product of its marginal probabilities is $p{e}^{-p}{p}^2/2>0$.

At the suggested value $r=e^{-r}$, the covariance is zero, but the same conditional probability contradiction still applies. This is an explicit example of how uncorrelated Bernoulli and Poisson variables need not be independent. Zero covariance controls one mixed moment, whereas independence requires the entire joint probability mass function to factor.

The marginal distributions have means $\mathbb{E}[X]=p$ and $\mathbb{E}[Y]=p$. In the joint distribution, $XY$ is nonzero only at $(1,1)$: when $X=1$, the only possible values of $Y$ are $0$ and $1$. HenceThese quantities are finite; for example, $0⩽XY⩽1$. The covariance changes sign when $p=e^{-p}$.

Equality can occur only at the two pairs $(0,0)$ and $(1,1)$. Their combined probability isTaking the complementary probability givesSince $1+2p>0$, the supplied exponential bound impliesThis is a bound on the mismatch probability for this particular coupling of the two marginal distributions; the variables are not presumed independent.

Summing the two nonzero entries with $X=1$ givesThe remaining probability is $\mathbb{P}(X=0)=1-p$, so $X$ has the Bernoulli distribution with parameter $p$.

For the other marginal, the two entries at $Y=0$ add to $e^{-p}$, the entry at $Y=1$ is $p{e}^{-p}$, and for $k⩾2$ the only nonzero entry is already $e^{-p}{p}^k/k!$. Thusand $Y$ has the Poisson distribution with parameter $p$. This also checks normalization of the joint probability mass function by summing the marginal.

Interpret existence of the expectations as finite expected values, equivalently $\mathbb{E}|X|<\infty$ and $\mathbb{E}|Y|<\infty$. First justify the summations rather than assuming that a signed double series can be rearranged. Independence and Tonelli theorem giveConsequently $XY$ is an integrable random variable, and Fubini theorem permits summing the signed terms in either order:If extended infinite expectations are allowed instead, the phrase “exist” needs additional care: expressions such as $0\cdot \infty$ do not define an expectation. The finite-expectation convention avoids that ambiguity.

The random variables are independent random variables if the events determined by one are independent of the events determined by the other. For integer-valued variables, this is equivalent to the factorizationIndeed, independence gives the factorization by taking singleton events. Conversely, summing it over $x\in A$ and $y\in B$, for any subsets $A,B\subseteq \mathbb{Z}$, gives $\mathbb{P}(X\in A,Y\in B)=\mathbb{P}(X\in A)\mathbb{P}(Y\in B)$. Thus the singleton condition implies independence of all events in question.

For each integer $x$, the events $\{X=x,Y=y\}$, indexed by the integers $y$, are disjoint and their union is $\{X=x\}$. Therefore countable additivity gives the marginal probability mass functionThe nonnegative terms make this sum well-defined, even when the support is infinite.

Use mathematical induction on the iteration. Initially $x_0=b=1\cdot b$. If $x_n={\lambda }_{n}b$, then $A{x}_n={\lambda }_{n}b/2$, $x_n^{T}b={\lambda }_n$, and $x_n^{T}A{x}_n={\lambda }_n^2/2$. ThereforeA gradient descent step with any scalar step length ${\alpha }_n$ consequently has the formThus every iterate remains on the same eigenline, withThis proves that gradient descent preserves an eigenline for a cubic product. The statement is invariance, not convergence to a minimum: along the negative direction of this line the cubic polynomial objective is unbounded below.

At $x=0$, the gradient from the preceding part vanishes and the Hessian matrix is zero. The second-order test is therefore inconclusive. Instead inspect the eigenline:For arbitrarily small positive and negative $t$, these values have opposite signs. Thus $0$ is neither a local maximum nor a local minimum. In dimension at least two it is a degenerate saddle point; in one dimension it is a stationary inflection point. Looking only at the zero Hessian matrix would not determine this classification.

Put $q(x)=x^{T}Ax$ and $t(x)=x^{T}b$. The product rule givesDifferentiate each term. The derivative of $2tAx$ is $2(Ax)b^T+2tA$, while that of $qb$ is $2b(Ax{)}^T$. HenceThe last two terms are transposes of each other, so this Hessian matrix is symmetric, as expected.

The matrix $C=A-b{b}^T$ is an invertible matrix. Indeed, on the line spanned by $b$ it has eigenvalue $-1/2$, and on $b^{\perp }$ it agrees with the positive-definite restriction of $A$. Therefore the full Newton step is well defined and givesThis is the unique stationary point because $C$ is an invertible matrix. The algebra applies to every starting point, but it does not make the answer a minimum: for $d\ge 2$ it is the saddle point just classified, and for $d=1$ it is a maximum. A safeguarded minimization algorithm that replaces an indefinite matrix used as its Hessian matrix or damps the step need not take this exact step; the assertion concerns the ordinary full Newton step formula.

The gradient vanishes at $0$. Decompose any vector as $x=tb+w$, with $w\perp b$. Symmetry of $A$ and $Ab=\frac{1}{2}b$ imply $b^{T}Aw=(Ab{)}^{T}w=0$. Since $\Vert b\Vert =1$,If $d\ge 2$, choose a nonzero $w\perp b$. Arbitrarily close to zero, the direction $b$ gives negative values and the direction $w$ gives positive values. Thus $0$ is a saddle point for $d\ge 2$. Equivalently, the Hessian matrix has one negative eigenvalue and all its other eigenvalues positive.

The dimension matters: if $d=1$, the orthogonal complement is zero-dimensional and $g(tb)=-t^2/2$. In that case $0$ is the unique global maximum. This edge case is not excluded by the stated hypotheses. The decomposition is a rank-one downdate along an eigenvector.

Write $(x^{T}b{)}^2=x^{T}b{b}^{T}x$ and put $C=A-b{b}^T$, which is symmetric. Differentiating this quadratic form givesUnlike $A$, $C$ need not be a positive-definite matrix. Its behaviour must be checked before classifying the stationary point.

The full, undamped Newton's method in optimization step isThe eigenvector identity gives $A^{-1}b=2b$. Consequently $x_1=-b$ for every starting point. This is a one-step Newton solution of a quadratic stationary equation: its gradient is an affine function, so a single exact Newton solve resolves it completely.

The eigenvector identity $Ab=\frac{1}{2}b$ gives $\nabla f(-b)=-2Ab+b=0$. Since $H_f=2A$ is a positive-definite matrix, this is a strict local minimum. A global certificate follows by completing the quadratic form:Positive definiteness makes the first term nonnegative, with equality only at $x=-b$. Thus $-b$ is the unique global minimum, with value $-1/2$.

Use compatible dimensions $A\in {\mathbb{R}}^{d\times d}$ and $b,x\in {\mathbb{R}}^d$; the printed “b-by-b” size is read consistently with the displayed products and domain. Take derivatives as column gradients. Symmetry of $A$ givesFor example, the directional differential of $x^{T}Ax$ is $d{x}^{T}Ax+x^{T}Adx=2(Ax{)}^{T}dx$. The linear term contributes $b$ to the gradient and zero to the Hessian matrix.