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The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}(e^{tX})$, defined for real $t$ where that expectation is finite. For a standard normal distribution, completing the square givesIt is finite for every real $t$. Derivatives at zero give the moments: the expansion $e^{t^2/2}=1+t^2/2+t^4/8+O(t^6)$ therefore givesFor $Y$ with the normal distribution of mean $\mu$ and variance ${\sigma }^2$, write $Y=\mu +\sigma X$ in distribution. Its moment-generating function is consequentlyThis formula also covers the constant random variable when $\sigma =0$.

For $0\le u\le 1$, independence givesThus $U$ has probability density function $f_U(u)=2(1-u)$, and its expected value isPointwise, $U+V=X+Y$ and $UV=XY$. By linearity of expectation, $\mathbb{E}V=1-1/3=2/3$; by independence, $\mathbb{E}(UV)=\mathbb{E}X\mathbb{E}Y=1/4$. Consequently the covariance of two uniform order statistics isThe order statistics are dependent despite the independence of the original observations.

The differential operator is $(D-3{)}^2$, where $D=d/dx$. Factoring out the forcing's exponential removes this repeated characteristic root of a constant-coefficient differential equation: set $y=e^{3x}v$. Differentiation gives $(D-3)y=e^{3x}{v}^{\prime }$ and $(D-3{)}^{2}y=e^{3x}{v}^{\prime \prime }$, so $v^{\prime \prime }=\cos (2x)$. Two integrations yieldThe initial conditions become $v(0)=0$ and $v^{\prime }(0)+3v(0)=1$, hence $B=1/4$ and $A=1$. ThusThe factorization verifies the differential equation directly, and the displayed expression has the required value and derivative at zero.

Let $\Theta =\angle AOB$ and let the sphere have radius $R$. By rotational invariance, condition on $A$ and place it at the north pole. A band at polar angle $\theta$ has surface area $2\pi R^2\sin \theta d\theta$, while the sphere has area $4\pi R^2$. Thus the central angle between independent uniform sphere points has densitywith density zero outside that interval.

Condition now on $A,B$ with their angle fixed at $\theta$. For $\angle AOC$ and $\angle BOC$ to be acute, the unit position vector of $C$ must have positive dot products with both unit vectors of $A$ and $B$. Hence $C$ must lie in the intersection of the two corresponding open hemispheres. The great circles bounding them meet at the two antipodes perpendicular to the plane of $A,B$. Their common hemisphere region is a spherical lune of opening $\pi -\theta$: when $A=B$ it is a hemisphere of opening $\pi$, and as the two normals separate, that opening decreases by $\theta$. More explicitly, take the common boundary antipodes as poles. The longitude conditions each allow an interval of length $\pi$, with their centres separated by $\theta$, so the intersection has longitude width $\pi -\theta$.

Integrating the spherical area element $R^2\sin \phi d\phi d\lambda$ over that lune gives area $2R^2(\pi -\theta )$. Thus the conditional probability that the two angles involving $C$ are acute is $(\pi -\theta )/(2\pi )$. The remaining angle $\Theta$ must itself be acute, so the desired probability isIntegration by parts gives ${\int }_0^{\pi /2}\theta \sin \theta d\theta =1$, while ${\int }_0^{\pi /2}\sin \theta d\theta =1$. ThereforeThis calculates pairwise acute central angles of three uniform sphere points by conditional area, rather than incorrectly treating the three angle events as independent. Boundary right angles have probability zero, so choosing open rather than closed hemispheres does not change the result.

The inverse change of variables is $x=u$, $y=(v-u)/a$, with absolute Jacobian determinant $1/a$. Because the original random variables are independent, their joint density is $f(x)g(y)$. Thus the transformed joint density isThe support is important: $X,Y\ge 0$ implies $V\ge U\ge 0$.

For independent exponential random variables of rate $\lambda >0$ and $a=1/2$, this becomes $2{\lambda }^2{e}^{-2\lambda v+\lambda u}$ on $0\le u\le v$. Integrating out $u$ gives the density of the sum:It vanishes for $v<0$. On the other hand, the cumulative distribution function of the maximum isDifferentiating yields exactly the same density. The maximum and $X+Y/2$ have the same distribution, as described by maximum of two exponentials as a scaled sum; this does not assert that they are equal for the same pair of sampled values.

Let $F(x)=\sqrt{2+\sqrt{x}}$ for $x\ge 0$. This is an increasing continuous function. Starting at $a_1=\sqrt{2}$, induction gives $\sqrt{2}\le a_n\le 2$: if $a_n\le 2$, then $a_{n+1}\le \sqrt{2+\sqrt{2}}<2$, and every iterate is at least $\sqrt{2}$.

The first step has $a_2>\sqrt{2}=a_1$. Since $F$ is increasing, $a_n\ge a_{n-1}$ implies $a_{n+1}=F(a_n)\ge F(a_{n-1})=a_n$. Thus this sequence is increasing and bounded above. Let $L_{-}={\sup }_n{a}_n$. For every $\epsilon >0$, a term exceeds $L_{-}-\epsilon$; all later terms lie between it and $L_{-}$. This proves convergence, the elementary bounded monotone sequence theorem.

For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt{2}$ because every positive iterate is at least that value. Hence it converges to its infimum $L_{+}$. From $b_2=2$ onward all terms lie in $[\sqrt{2},2]$.

Continuity of $F$ permits passing to each limit in the recurrence, giving $L_{\pm }^2=2+\sqrt{L_{\pm }}$. To prove that these fixed points agree, suppose $r,s\ge \sqrt{2}$ both satisfy that equality. Subtracting the two equations and rationalizing givesThe second factor is positive: $r+s\ge 2\sqrt{2}$, whereas $(\sqrt{r}+\sqrt{s}{)}^{-1}\le 1/(2\sqrt{\sqrt{2}})<2\sqrt{2}$. Thus $r=s$. ThereforeExistence has been supplied by the two convergent sequences, and the subtraction argument gives the required uniqueness in their common interval.