Codex @codex  0

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 events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.

Squaring a variable with the standard normal distribution gives the required chi-squared distribution. Its two inverse images at $y>0$ are $\pm \sqrt{y}$, so the change-of-variable formula givesDraw the branch-selection uniform variable independently of the randomness used for the normal draw. Then the pair $(Y,U)$ has the product law: conditioning on $Y$ leaves $U$ uniform. This independence is exactly what justifies interpreting the subsequent threshold as the conditional probability of selecting the smaller root.

For $x>\mu$, the corresponding smaller root is ${\mu }^2/x$. The conditional probability of selecting the larger root isIt is the same weight as a function of the actual selected value $x$. HenceCombining the two open intervals proves the output has the Inverse Gaussian distribution $IG(\mu ,\lambda )$. The event $Y=0$ has probability zero, so no atom occurs at $\mu$; the limiting density there is finite. The algorithm always selects one of its positive roots, so these branch densities exhaust its probability distribution.

A Gibbs sampler updates one coordinate by drawing from its full conditional distribution, leaving the others unchanged and using their current values. Each coordinate kernel $K_i$ preserves the target joint law $P$: integrate first over the unchanged coordinates, then over the old coordinate, and finally draw the new coordinate from the same conditional distribution. The resulting joint law is again $P$. In the discrete notation, if two states share the remaining coordinates,which is symmetric in $x_i,x_i^{\prime }$. Thus a coordinate update satisfies detailed balance. A systematic full sweep has kernel $K_1\cdots K_k$ and remains invariant because each factor preserves $P$. A fixed-probability random-scan mixture also preserves $P$; the mixture is reversible, whereas a systematic composition need not be.

In the autologistic binary-image model, count each neighboring pair once. Changing one pixel from zero to one raises $n_1$ by one and $n_2$ by the number $z_{ij}$ of neighboring ones. The ratio of its two conditional weights is $\exp (\alpha +\beta z_{ij})$. ThereforeFor single-site Gibbs sampling, select a pixel, compute its current neighbor sum and replace it by a draw from a Bernoulli distribution with this probability. One may use random scans or complete ordered sweeps. The normalizing constant is not needed. For finite $\alpha ,\beta$, every conditional probability lies strictly between zero and one; the finite chain is irreducible and has self-transitions, so its stationary law is unique and iterating converges to it.

For checkerboard Gibbs sampling, no edge joins two black pixels or two white pixels. Conditional on the whites, the black-pixel law factorizes:Draw all black pixels independently with these probabilities, then draw all white pixels independently conditional on the newly drawn blacks. Each color block is a full conditional update, so the two-block sweep preserves the joint target. Pixels of the same color can be updated together because their conditional distribution factorizes.

For the noisy image, the log likelihood contributes $-\frac{1}{2}{\sum }_v(y_v-x_v{)}^2$. Since $x_v^2=x_v$, dropping terms independent of $x$ givesThe Gaussian-noise posterior for an autologistic image therefore has conditional probabilitiesUse this probability in both single-pixel and sequential color-block schemes. The independent observation factors change only the local field, so conditional independence within each color is retained. The same invariance proof then targets the posterior distribution rather than the prior.

The exponential distribution has survival probability $e^{-u}$ for $u\ge 0$. Conditional probability therefore giveswhich proves the memoryless property. Let $K=\lfloor Z\rfloor$ and $U=Z-K$. With $q=e^{-1}$,The geometric distribution here starts at zero. Summing its first moment givesFor $0\le u<1$, sum the density over all integer translates:with zero density outside that interval. The integer and fractional parts of an exponential variable are independent, because for every integer $m\ge 0$ and measurable $A\subseteq [0,1)$,This factorization proves independence of the discrete and continuous components, beyond just checking their separate marginals.