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If $X$ and $Y$ are independent random variables with Gamma distributions of shapes $a,b>0$ and the same rate $\theta >0$, then $U=X/(X+Y)$ has the Beta distribution with parameters $(a,b)$, and $V=X+Y$ has the Gamma distribution of shape $a+b$ and rate $\theta$. Moreover $U$ and $V$ are independent. The inverse change of variables $(x,y)=(uv,(1-u)v)$ has absolute Jacobian determinant $v$, which makes the joint probability density factorize on the product domain $0<u<1$, $v>0$. The common rate is essential for this factorization.

Where the marginal density $f_X(x)$ is positive, a joint probability density determines the conditional densityIt integrates to one in $y$, and supplies conditional expectations by integrating against it. Its values on a probability-zero set of conditioning points are not fixed by the joint probability distribution.

For $n$ independent and identically distributed rate-$\lambda$ exponential random variables, let $M_0=0$ and $M_1<\cdots <M_n$ be their order statistics. The gaps $D_i=M_i-M_{i-1}$ are independent and have rates $(n-i+1)\lambda$. After the first clock rings, the memoryless property makes the surviving clocks fresh independent exponentials, independently of the elapsed minimum and the winning label. Induction gives the result. Equivalently, the ordered joint probability density is $n!{\lambda }^n\exp [-\lambda {\sum }_i{m}_i]$ on $0<m_1<\cdots <m_n$; changing to gaps gives a product of exponential densities because ${\sum }_i{m}_i={\sum }_i(n-i+1)D_i$.

For two independent rate-$\lambda >0$ exponential distributions, set $U=\min (X,Y)$ and $V=\max (X,Y)$. Summing the two ordering branches gives joint probability density $2{\lambda }^2{e}^{-\lambda (2u+d)}$ for $(U,d=V-U)$ with $u,d>0$. It factors into independent exponential probability density functions. Thus $V$ is distributed as $X+Y/2$, with expected value $3/(2\lambda )$ and variance $5/(4{\lambda }^2)$. This spacing result is the order-statistic manifestation of the memorylessness of the exponential distribution.

Let $M_t={\sup }_{0\le s\le t}{B}_s$ for a standard Brownian motion. For $m\ge 0$, the Brownian reflection principle givesOn $b<m$, the pair $(B_t,M_t)$ therefore has joint probability density

A marginal density is a probability density function for one component of a joint probability distribution. Integrating the joint probability density over the other coordinates gives it; for two components, $f_X(x)=\int f_{X,Y}(x,y)dy$.

For a fixed $x\in [0,1]$, the allowed $y$ interval is $[0,\sqrt{x}]$. Integrating the joint probability density over it gives the marginal densityFor a fixed $y\in [0,1]$, the inequality $y\le \sqrt{x}$ is equivalent to $x\ge y^2$. The other marginal density is thereforeAs a normalization check, ${\int }_0^{1}3x^{2}dx=1$ and ${\int }_0^{1}3y(1-y^4)dy=1$.

The triangular support can be written $0\le x\le \pi /2$, $\pi /2-x\le y\le \pi /2$. The density is nonnegative there, and the normalization of a probability distribution requiresHence $\boxed{A=2}$. Multiplying the joint probability density by the required function cancels $\sin x$ on the interior. Its expectation isThe singularity at $x=0$ lies at a support vertex of zero probability; the displayed integral establishes integrability. Assigning any value to the random function at that single vertex does not change the expectation.

Take $\lambda >0$ as the rate of the exponential distribution. By independence, the joint probability density of the original pair is ${\lambda }^2{e}^{-\lambda (x_1+x_2)}$ on $x_1,x_2>0$. Put $s=Y_1$, $r=Y_2$. The inverse change of variables isIts Jacobian determinant isUsing the absolute value in the change of variables formula, the transformed joint probability density isand zero otherwise. The support is the entire positive quadrant, and the factors are separately normalized:Integrating out either variable therefore yields these factors as the marginal densities. Factorization of the joint probability density proves independence, withThe sum has a shape-two gamma distribution, and the ratio has a beta-prime distribution with parameters one and one, independent of $\lambda$. Its cumulative distribution function is $r/(1+r)$ for $r>0$.

A marginal statistical likelihood uses a statistic whose marginal sampling distribution depends on the parameter of interest but does not involve the nuisance parameter. It integrates out other observed coordinates. A profile likelihood instead keeps the full data and, for every fixed interest parameter, fits the nuisance parameter by maximizing the full likelihood. Profiling supplies a likelihood function but usually not a normalized sampling probability density function. Here the marginalization is of data coordinates; Bayesian model evidence averages parameters against a prior and is a separate construction.

Use shape-rate notation for the gamma distribution. The independent sums have $X\sim \mathrm{Gamma}(m,\psi \lambda )$ and $Y\sim \mathrm{Gamma}(n,\lambda )$, so their joint probability density for $x,y>0$ isFor $x=tu$, $y=u$ the absolute Jacobian determinant is u. ThereforeIntegrate u using the gamma integral ${\int }_0^{\infty }{u}^{m+n-1}{e}^{-bu}du=\Gamma (m+n)b^{-(m+n)}$. The lambda factors cancel, leavingThus T gives a nuisance-free marginal likelihood. Discarding only terms independent of psi, its logarithm isThe factor t in the logarithm is required, as in the original PDF.

The full log-likelihood is $m\log \psi +(m+n)\log \lambda -\lambda (\psi x+y)+c(x,y)$. Its lambda derivative vanishes uniquely atThe second derivative in lambda is negative. Substituting this maximum gives ${\ell }_p(\psi )=m\log \psi -(m+n)\log (\psi x+y)+c^{\prime }(x,y)$. Since $\psi x+y=y(1+\psi t)$, it agrees with ${\ell }_m(\psi ;t)$ up to a psi-independent constant. This proves the Gamma-ratio marginal and profile likelihood identity. Their common maximizer is $\hat{\psi }=m/(nt)=my/(nx)$, if an interest-parameter estimate is also wanted.

All the following distributions are conditional on the point lying in the unit disk. Initially the joint probability density of $(S,T)$ is $1/4$ on the square. The disk has area $\pi$, so the conditioning event has probability $\pi /4$. Dividing by this probability gives the constant conditional density $1/\pi$ on the disk, proving its uniform distribution there.

The polar coordinates $S=r\cos \theta ,T=r\sin \theta$ have Jacobian determinant $r$. ConsequentlyIntegrating the joint probability density gives $f_R(r)=2r$ and $f_{\Theta }(\theta )=1/(2\pi )$. Since the joint density equals the product of these marginal densities on a product support, $R$ and $\Theta$ are independent. In particular $U=R^2$ has the uniform distribution on $(0,1)$.

The second change of variables preserves the angle and changes the radius toThe event $R=0$ has probability zero, so arbitrary definitions at that point have no effect on the distribution. For $q\ge 0$, monotonicity of the logarithm givesSince $Q$ is a function of $R$, it remains independent of the uniform angle. Hence the new joint probability density isFinally, $(X,Y)=(Q\cos \Psi ,Q\sin \Psi )$ has polar Jacobian determinant $q$. Dividing the preceding density by $q$ gives the Cartesian joint probability densityThis factors into two standard normal distribution densities on ${\mathbb{R}}^2$. Thus $X$ and $Y$ are independent, each with mean zero and variance one. This derivation is the radial form of the Box-Muller transform.

Let $Q=P_1-P_0$ and $R=I-P_1$. For nested subspaces these are orthogonal projections onto ${\Theta }_1\cap {\Theta }_0^{\perp }$ and ${\Theta }_1^{\perp }$, of ranks $d_1-d_0$ and $d-d_1$. Under the null, both annihilate $\mu$. In an orthonormal basis adapted to ${\Theta }_0$, these two spaces and their complement, the coordinates of $X-\mu$ are independent $N(0,{\sigma }^2)$. Consequentlyand the two variables are independent. Equivalently the joint probability density of the two unscaled sums $a,b>0$ factors into gamma densities of respective shapes $(d-d_1)/2,(d_1-d_0)/2$ and common scale $2{\sigma }^2$.

The nuisance variance cancels inReject for a value above the upper $\alpha$ critical quantile to obtain a level-$\alpha$ test. Under a genuine alternative the numerator is noncentral chi-square with noncentrality $\Vert Q\mu {\Vert }^2/{\sigma }^2$, while the denominator retains its central law, explaining the upper-tail choice.

Set $\mathbf{e}=1/\sqrt{n}$ and $\mathbf{f}=\mathbf{x}/\sqrt{S_{xx}}$. These are orthonormal because ${\sum }_i{x}_i=0$. Complete them to an orthonormal basis $\mathbf{e},\mathbf{f},{\mathbf{q}}_3,\dots ,{\mathbf{q}}_n$ of ${\mathbb{R}}^n$. An orthogonal transformation of the standard normal error vector $\mathbf{\epsilon }/\sigma$ has independent normal random variables as coordinates $Z_1,\dots ,Z_n$: its joint probability density remains $(2\pi {)}^{-n/2}\exp (-\Vert z{\Vert }^2/2)$ because the transformation preserves squared length and has Jacobian determinant of absolute value one.

Projection onto these basis directions givesTherefore the joint distribution of least-squares and variance estimators isThe last law is the chi-squared distribution of a sum of $n-2$ squared independent normal random variables with mean zero and variance one, and independence follows because the three statistics use disjoint independent coordinate sets. When all $x_i=0$, omit the slope direction: $\hat{\alpha }$ and $n{\hat{\sigma }}^2/{\sigma }^2\sim {\chi }_{n-1}^2$ are independent, and there is no identifiable slope estimator.

The exponential distribution has no atoms, so all sample values are distinct almost surely, and $M_i$ is the $i$th order statistic. Set $M_0=0$ and $D_i=M_i-M_{i-1}$. Part (b) gives $D_1\sim \mathrm{Exp}(n\lambda )$.

To justify induction and the required independence, let $J$ be the label attaining the minimum. Conditional on its time and label, the remaining clocks are fresh independent rate-$\lambda$ exponential random variables, by the memoryless property. This conditional claim can be checked directly: for a specified label $j$, first ringing time $u$, and residual lifetimes $y_k>0$ for $k\ne j$, the joint probability density isThus the first minimum, its uniform winning label and the surviving residual clocks are independent. Repeat this argument after each minimum. At the $i$th step there are $n-i+1$ clocks, so the independent exponential order-statistic spacings satisfySince $Y_k/k$ has rate $k\lambda$ whenever $Y_k\sim \mathrm{Exp}(\lambda )$, summing the gaps provesIn particular the maximum is $M^{\ast }=M_n$. By variance additivity for independent random variables and the Laplace transform of an exponential distribution, its requested quantities areThe product follows from independence of the gaps, not merely from their marginal probability distributions.

The independent uniform distributions give a constant joint probability density $1$ on the unit square. For $0<x<1$, integrate over $x<u<v^2$:For the random quadratic with uniform coefficients, real roots require its discriminant to be nonnegative, equivalently $U\le V^2$. The boundary has probability zero, and the area under this parabola isOn that event the roots are $-V\pm \sqrt{V^2-U}$, both nonpositive. Bounding both absolute values by one therefore amounts to bounding the more negative root:Since $0\le V\le 1$, squaring the inequality is legitimate and gives $U\ge 2V-1$. The admissible unit-square region is $\max (0,2v-1)\le u\le v^2$. Its probability isDividing by the real-root probability gives the requested conditional probability:

For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

The independent and identically distributed random variables have a continuous probability distribution, so any tie has probability zero. Their joint probability density is unchanged by a permutation of their indices: this is order symmetry of independent random variables. The $n!$ strict orderings are disjoint and exhaust the sample space apart from the zero-probability ties. Each ordering therefore has the same probability, and their probabilities sum to one.

Set $U=X/(X+Y)$ and $V=X+Y$. Their inverse change of variables is $X=UV$, $Y=(1-U)V$, with domain $0<U<1$, $V>0$. The absolute Jacobian determinant of this inverse transformation isUse the independence of random variables to multiply the two original Gamma distribution densities. The change of variables formula then gives the joint probability densityFactor this into two normalized densities:The support is a product domain, and the factors are precisely a Beta distribution density and a Gamma distribution density. ThereforeThis beta-gamma independence uses the common rate $\theta$; it is a rate parameter, not the scale parameter $1/\theta$.

Integrate the joint probability density over $y$ to obtain the marginal density of $X$:Thus $X$ has an exponential distribution of rate one. Dividing the joint probability density by this marginal density gives the conditional densityEquivalently, $Y\mid X=x$ has an exponential distribution of rate $x$. The value at $x=0$ can be assigned arbitrarily: it is a probability-zero conditioning value and the density ratio there is not meaningful.