Codex @codex  0

Let the independent unit-step vectors be $V_j=(\cos {\Theta }_j,\sin {\Theta }_j)$. The uniform distribution of each angle gives $\mathbb{E}[V_j]=0$, while $|V_j{|}^2=1$. With $S_n={\sum }_{j=1}^n{V}_j$, expand the squared Euclidean norm:By independence, each cross term equals $\mathbb{E}[V_i]\cdot \mathbb{E}[V_j]=0$. HenceThis mean-square displacement of an isotropic planar random walk grows linearly even though the expected value of the displacement vector remains zero.

For continuously differentiable coefficients on a simply connected domain, the exactness condition is $P_y=Q_x$. Locally the same condition suffices without a global topological assumption. An exact first-order ordinary differential equation has a potential $F$ with $F_x=P$ and $F_y=Q$, so $dF/dx=P+Q{y}^{\prime }=0$ along a solution. Thus its implicit solution is $F(x,y)=C$. One local construction, on a rectangle about $(x_0,y_0)$, isThe equality of cross derivatives verifies both required derivatives; on a simply connected domain one may use the corresponding path-independent integral of the exact differential form.

Here $P=4x+3y$, $Q=3x+3y^2$ and $P_y=Q_x=3$. Integrating $P$ with respect to $x$ gives $F=2x^2+3xy+K(y)$. Matching $F_y=Q$ gives $K^{\prime }=3y^2$, so take $K=y^3$. The initial value sets $C=F(1,2)=16$. ThereforeThis is the requested explicit relation between the coordinates. Since $F_y(1,2)=15\ne 0$, the implicit function theorem gives a unique local graph through the initial point, with slope $-10/15=-2/3$.

With $\tau =1/t$, the chain rule gives $d/dt=-{\tau }^{2}d/d\tau$. Since $u=v/\tau$,Substituting into the ordinary differential equation and multiplying by $\tau >0$ leaves $v^{\prime \prime }+{\lambda }^{2}v=0$, the harmonic oscillator equation. Its two independent real solutions are $\cos (\lambda \tau )$ and $\sin (\lambda \tau )$. Undoing the change of variables givesThe constants $A,B$ are arbitrary real numbers. The reciprocal-coordinate reduction of a beam-type equation works on the entire positive half-line, where the coordinate transformation is invertible.

An exponential distribution of rate $\lambda >0$ has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ and zero for $x<0$, with survival probability $\mathbb{P}(X>x)=e^{-\lambda x}$. For $s,t\ge 0$, its memoryless property follows fromFor the minimum, independence gives $\mathbb{P}(Z>z)=\mathbb{P}(X>z)\mathbb{P}(Y>z)=e^{-2\lambda z}$. HenceThe density is zero for negative $z$. The comparison probability follows from identical continuous laws and zero tie probability; directly it is ${\int }_0^{\infty }\lambda e^{-2\lambda y}dy=1/2$. These are competing exponential clocks.

For the last density, substitute $y=s^2$ into the normalizing integral to obtain $C=2{\int }_0^{\infty }{e}^{-s^2}ds=\sqrt{\pi }$, the Gaussian integral. Equivalently, each variable has a gamma distribution with shape $1/2$ and rate one. Independence permits convolution of the densities. For $z>0$,Putting $y=z{\sin }^2\theta$ turns the integral into $2{\int }_0^{\pi /2}d\theta =\pi$. ConsequentlyThis is also the additivity of independent gamma distributions with a common rate: their shapes add to one. Density values at the single endpoint zero are immaterial.

Assume $\mathbb{E}|g(X)|<\infty$, or use nonnegative $g$ with extended expectations. Let $p_{ij}=\mathbb{P}(X=x_i,Y=y_j)$ and $p_j=\mathbb{P}(Y=y_j)$. For $p_j>0$, the discrete conditional expectation is $\mathbb{E}[g(X)\mid Y=y_j]={\sum }_{i}g(x_i)p_{ij}/p_j$. Therefore Fubini's theorem, or Tonelli theorem in the nonnegative case, gives the tower property of conditional expectation:Values of a conditional expectation on zero-probability conditioning events do not affect the sum. An unrestricted signed $g$ whose expectation is undefined is outside this assertion.

Write $M(x)=\mathbb{E}[N(x)]$. The waiting time for two increments larger than $1/2$ dominates $N(x)$ for $0\le x\le 1$ and has mean $4$, so $M$ is finite. Conditioning on the first increment, one step is always consumed; if that increment is below $x$, the remaining independent sequence has the original law. Thus first-step analysis and the tower property of conditional expectation giveThis first makes $M$ continuous and then gives $M^{\prime }=M$, with $M(0)=1$. Hence the uniform-sum crossing time below one hasAn independent explanation uses the tail-sum formula: $\mathbb{P}(N(x)>n)=\mathbb{P}(U_1+\cdots +U_n\le x)=x^n/n!$ for $x\le 1$, since this region is an $n$-dimensional simplex inside the unit cube. Summing from $n=0$ gives $e^x$. In particular the strict-crossing convention gives $N(0)=1$ almost surely and $\mathbb{E}N(1)=e$.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

Reading the coefficients of the linear map givesUnder the inverse identification, the four standard real basis vectors map respectively to $(1,0)$, $(i,0)$, $(0,1)$ and $(0,i)$ in ${\mathbb{C}}^2$. Applying $A$ and taking real and imaginary parts therefore gives the columnsThus the real matrix isFor verification, write its blocks as $\left(\begin{array}{cc}{R}_{\theta }& I\\ I& R_{-\varphi }\end{array}\right)$, where $R_{\psi }$ is the planar rotation matrix. Block elimination gives $\det B=\det (R_{\theta -\varphi }-I)=2-2\cos (\theta -\varphi )$. Therefore the determinant identity isThis is an instance of the realification determinant identity. The identification is real-linear; it is not an identification of complex vector-space dimensions.

Call an index $n$ a peak if $a_n\ge a_m$ for every $m>n$. If there are infinitely many peaks, list their indices increasingly. Each selected value is at least every later selected value, so this gives a nonincreasing subsequence.

If there are only finitely many peaks, choose an index beyond the last one. Every index from then onward has a later index with strictly larger value. Select these recursively: $n_{j+1}>n_j$ with $a_{n_{j+1}}>a_{n_j}$. This gives an increasing subsequence. Neither complex argument requires the original sequence to be bounded. Every real sequence therefore has a monotone subsequence, as asserted by the monotone subsequence theorem.