
Generating Samples
Owen, Chapters 3–6 and 15–17
Assignment 2 due Sep 11
Assignment 3 due Sep 25
Test 1 · Sep 15
October 9, 2026
\[ \vU_i \sim \Unif[0,1]^d \;\longrightarrow\; \vX_i \sim \varrho_{\mathrm{tar}} \;\longrightarrow\; Y_i = f(\vX_i) \]
\[ \mu=\Ex(Y),\qquad \hmu_n=\frac1n\sum_{i=0}^{n-1}Y_i \]

Uniform numbers are the raw material
A transformation produces the target distribution
A sampling design supplies
\[ U_i\sim\Unif[0,1],\qquad i=0,1,\ldots \]
The design may be IID, Latin hypercube, low discrepancy, or something else
For a target cumulative distribution function \(F\), define
\[ Q(u):=\inf\{x\in\reals:F(x)\ge u\} \]
Then \(Q(u)\le x \iff u\le F(x)\)
For \(X:=Q(U)\)
\[ \Prob(X\le x)=\Prob\{Q(U)\le x\} =\Prob\{U\le F(x)\}=F(x) \]
Thus each \(X_i:=Q(U_i)\) has the target CDF \(F\)
IID inputs yield IID targets
Other designs (Latin hypercube, low discrepancy, …) retain their dependence structure
Try the quantile-transform examples in the companion GeneratingSamples notebook
Suppose \(X\sim\Bin(3,0.6)\)
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| PMF \(\varrho_X(x)\) | \(0.4^3\) \(=0.064\) |
\(3(0.6)(0.4^2)\) \(=0.288\) |
\(3(0.6^2)(0.4)\) \(=0.432\) |
\(0.6^3\) \(=0.216\) |
| \(x\in\) | \([0,1)\) | \([1,2)\) | \([2,3)\) | \([3,\infty)\) |
| CDF \(F(x)\) | \(0.064\) | \(0.352\) | \(0.784\) | \(1\) |
| \(u\in\) | \((0,0.064]\) | \((0.064,0.352]\) | \((0.352,0.784]\) | \((0.784,1]\) |
| Quantile \(Q(u)\) | \(0\) | \(1\) | \(2\) | \(3\) |
E.g., \(U_0=0.63\), \(U_1=0.02\), and \(U_2=0.47\) produce \(X_0=2\), \(X_1=0\), and \(X_2=2\)
\(\exstar\) If \(u=0.70\), what value does the inverse transform produce? What about \(u=0.90\)?

\(F\) is right-continuous; \(Q\) is left-continuous
Let \(X\) be nonnegative with
\[ \Prob(X=0)=p_0, \qquad X\mid X>0\sim\Exp(\lambda) \]
Its CDF is
\[ F(x)= \begin{cases} 0, & x<0,\\ p_0, & x=0,\\ p_0+(1-p_0)(1-\me^{-\lambda x}), & x>0 \end{cases} \]
For the plot, \(p_0=0.3\) and \(\lambda=1\)

Inverting the CDF gives
\[ Q(u)= \begin{cases} 0, & 0<u\le p_0,\\[0.4em] -\dfrac{1}{\lambda}\log\!\left(\dfrac{1-u}{1-p_0}\right), & p_0<u<1 \end{cases} \]
\(\exstar\) Verify that \(F\bigl(Q(u)\bigr)\ge u\) and identify where equality fails
For the plot, \(p_0=0.3\) and \(\lambda=1\)

\(\exstar\) Define \(T(u):=Q(1-u)\).
Simplify \(T\) and explain why \(T(U)\) has the desired distribution for \(U\sim\Unif(0,1)\) even though \(T\) is not the quantile function.
Why might \(T\) be preferred to \(Q\)?
The generalized inverse handles atoms and continuous pieces with one definition
Let the one-dimensional target be a Gaussian mixture:
\[ X\sim p\Norm(\mu_1,\sigma_1^2)+(1-p)\Norm(\mu_2,\sigma_2^2), \qquad 0<p<1 \]
Writing \(\phi\) and \(\Phi\) for the standard normal PDF and CDF,
\[\begin{align*} \varrho_X(x) &=\frac{p}{\sigma_1}\phi\!\left(\frac{x-\mu_1}{\sigma_1}\right) +\frac{1-p}{\sigma_2}\phi\!\left(\frac{x-\mu_2}{\sigma_2}\right),\\ F_X(x) &=p\Phi\!\left(\frac{x-\mu_1}{\sigma_1}\right) +(1-p)\Phi\!\left(\frac{x-\mu_2}{\sigma_2}\right) \end{align*}\]
The PDF and CDF have analytic formulas, but the mixture quantile \(Q=F_X^{-1}\) generally has no analytic closed form
Numerically inverting \(F_X\) would work, but a hierarchical construction avoids solving \(F_X(x)=u\)
Generate independent \(U_1,U_2\sim\Unif(0,1)\)
The last coordinate \(U_2\) selects the component; \(U_1\) supplies the normal quantile
\[ C=\begin{cases}1,&U_2\le p,\\2,&U_2>p,\end{cases} \qquad X=\mu_C+\sigma_C\Phi^{-1}(U_1) \]
Then \(\Prob(C=1)=p\) and \(X\mid C=k\sim\Norm(\mu_k,\sigma_k^2)\)
A mixture chains a discrete component choice with a within-component transform
Take
\[\begin{align*} p&=0.3,\\ (\mu_1,\sigma_1)&=(-2,0.5),\\ (\mu_2,\sigma_2)&=(1,1) \end{align*}\]
For \(U_1=0.90\) and \(U_2=0.24\),
\[\begin{align*} C&=1,\\ X&=-2+0.5\Phi^{-1}(0.90)\\ &\approx -1.36 \end{align*}\]
\(\exstar\) Use \(U_1=0.25\) and \(U_2=0.70\).
Which component is selected?
What value of \(X\) is generated?

For \(\vX=(X_1,\ldots,X_d)\) with independent marginals
\[ \vX=\bigl(Q_1(U_1),\ldots,Q_d(U_d)\bigr), \qquad \vU\sim\Unif[0,1]^d \]
where \(Q_j\) is the quantile function for the target distribution of \(X_j\)
Store \(n\) sampled vectors as an \(n\times d\) array:
\[ \mX= \begin{pmatrix} X_{01} & X_{02} & \cdots & X_{0d}\\ \vdots & \vdots & & \vdots\\ X_{n-1,1} & X_{n-1,2} & \cdots & X_{n-1,d} \end{pmatrix} \]
Rows are samples
Columns are coordinates
Set \(Z_j=\Phi^{-1}(U_j)\) for independent uniform inputs
Let \(\vZ\sim\Norm(\vzero,\mI_d)\) be a column vector of \(d\) independent standard normal random variables, so
\[ \Ex(\vZ)=\vzero, \qquad \Ex(\vZ\vZ^\mathsf{T})=\mI_d \]
The affine transport is \(\vT(\vz)=\mA\vz+\vmu\)
Define \(\vX=\vT(\vZ)\). Then
\[\begin{align*} \Ex(\vX) &= \vmu,\\ \cov(\vX) &= \Ex\bigl[(\vX-\vmu)(\vX-\vmu)^\mathsf{T}\bigr] =\Ex(\mA\vZ\vZ^\mathsf{T}\mA^\mathsf{T})\\ &=\mA\Ex(\vZ\vZ^\mathsf{T})\mA^\mathsf{T} =\mA\mA^\mathsf{T} \end{align*}\]
Every linear combination of the components of \(\vX\) is normal
Thus, if \(\mSigma=\mA\mA^\mathsf{T}\),
\[ \vX=\mA\vZ+\vmu\sim\Norm(\vmu,\mSigma) \]
For symmetric positive-definite \(\mSigma\), Cholesky factorization gives a lower-triangular \(\mA\) such that \(\mSigma=\mA\mA^\mathsf{T}\)
For
\[ \mSigma=\begin{pmatrix}3&-1\\-1&1\end{pmatrix}, \qquad \mA=\begin{pmatrix}a_{11}&0\\a_{21}&a_{22}\end{pmatrix} \]
matching the entries of \(\mSigma=\mA\mA^\mathsf{T}\) gives
\[\begin{align*} a_{11}^2&=3, &a_{11}a_{21}&=-1, &a_{21}^2+a_{22}^2&=1 \end{align*}\]
Hence
\[ a_{11}=\sqrt3, \qquad a_{21}=-\frac1{\sqrt3}, \qquad a_{22}=\sqrt{\frac23}, \qquad \mA_{\mathrm{Chol}} =\begin{pmatrix}\sqrt3&0\\-1/\sqrt3&\sqrt{2/3}\end{pmatrix} \]
For
\[ \mSigma= \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2\\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix}, \qquad \mA= \begin{pmatrix} \sigma_1 & 0\\ \rho\sigma_2 & \sigma_2\sqrt{1-\rho^2} \end{pmatrix} \]
Different factorizations generate the same distribution
They may behave differently under low discrepancy sampling
For the same covariance matrix as in the Cholesky example,
\[ \mSigma=\begin{pmatrix}3&-1\\-1&1\end{pmatrix}, \]
let \(c=\cos(\mpi/8)\) and \(s=\sin(\mpi/8)\)
The ordered eigendecomposition is
\[ \mV=(\vv_1\ \vv_2)= \begin{pmatrix}c&s\\-s&c\end{pmatrix}, \qquad \mLambda=\operatorname{diag}(2+\sqrt2,\,2-\sqrt2) \]
The PCA factor simplifies to
\[ \mA_{\mathrm{PCA}}=\mV\mLambda^{1/2} =\begin{pmatrix} 1+1/\sqrt2&1-1/\sqrt2\\ -1/\sqrt2&1/\sqrt2 \end{pmatrix} \]
Thus \(\mA_{\mathrm{PCA}}\mA_{\mathrm{PCA}}^\mathsf{T} =\mV\mLambda\mV^\mathsf{T}=\mSigma\), just as for the Cholesky factor
A Gaussian process is a random function whose values at any finite collection of inputs form a multivariate normal random vector
\[ G\sim\GP(m,K) \]
means
\[\begin{align*} \Ex[G(t)] &= m(t),\\ \Ex\bigl[\{G(t)-m(t)\}\{G(x)-m(x)\}\bigr] &= K(t,x) \end{align*}\]
For inputs \(t_1,\ldots,t_d\), let \(\mu_j=m(t_j)\)
\[ \bigl(G(t_1),\ldots,G(t_d)\bigr)^\mathsf{T} \sim\Norm(\vmu,\mSigma), \qquad \Sigma_{jk}=K(t_j,t_k) \]
Brownian motion \(B\) is the Gaussian process with
\[ m(t)=0, \qquad K(t,x)=\min(t,x), \qquad t,x\ge0 \]
For \(0=t_0<t_1<\cdots<t_d\)
\[ \bigl(B(t_1),\ldots,B(t_d)\bigr)^\mathsf{T} \sim\Norm(\vzero,\mSigma), \qquad \Sigma_{jk}=\min(t_j,t_k) \]
The covariance structure implies independent normal increments
For \(0<t_1<t_2<t_3\), let \(\Delta t_1=t_1\) and \(\Delta t_j=t_j-t_{j-1}\). Then
\[ \mSigma= \begin{pmatrix} t_1&t_1&t_1\\ t_1&t_2&t_2\\ t_1&t_2&t_3 \end{pmatrix}, \qquad \mA= \begin{pmatrix} \sqrt{\Delta t_1}&0&0\\ \sqrt{\Delta t_1}&\sqrt{\Delta t_2}&0\\ \sqrt{\Delta t_1}&\sqrt{\Delta t_2}&\sqrt{\Delta t_3} \end{pmatrix}, \qquad \mSigma=\mA\mA^\mathsf{T} \]
Each entry sums the increments shared by both times:
\[ (\mA\mA^\mathsf{T})_{jk} =\sum_{\ell=1}^{\min(j,k)}\Delta t_\ell =t_{\min(j,k)} \]
In general, \(A_{jk}=\sqrt{\Delta t_k}\) for \(k\le j\) and \(A_{jk}=0\) otherwise
Let \(Z_j=\Phi^{-1}(U_j)\), so \(Z_1,\ldots,Z_d\ \IIDsim\ \Norm(0,1)\)
Start from \(B(t_0)=0\)
\[ B(t_j)=B(t_{j-1})+\sqrt{t_j-t_{j-1}}\,Z_j, \qquad j=1,\ldots,d \]
Equivalently
\[ B(t_j)=\sum_{k=1}^j\sqrt{t_k-t_{k-1}}\,Z_k \]
The Cholesky construction is the familiar cumulative-sum construction for Brownian motion
For \(S_0>0\), drift \(\gamma\), and volatility \(\sigma>0\), define
\[ S(t)=S_0\exp\!\left[ \left(\gamma-\frac{\sigma^2}{2}\right)t+\sigma B(t) \right] \]
Then \(S(t)>0\) and its multiplicative increments are lognormal
\(\exstar\) Using \(B(t)\sim\Norm(0,t)\) and the normal moment-generating function, show that \(\Ex[S(t)]=S_0\me^{\gamma t}\).
Hence determine \(\gamma\) so that \(\Ex[S(t)]=S_0\me^{rt}\).
The correction \(-\sigma^2/2\) makes \(\gamma\) the mean growth rate
The median growth rate is \(\gamma-\sigma^2/2\)
Let
\[ m(t)=S_0\me^{(\gamma-\sigma^2/2)t}, \]
the median of \(S(t)\)
\(\exstar\) The normal density weights the shocks \(b\) and \(-b\) equally. Show that \[ \frac{m(t)\me^{\sigma b}+m(t)\me^{-\sigma b}}{2} =m(t)\cosh(\sigma b)>m(t),\qquad b\ne0. \] Why do equal and opposite log shocks not cancel in price?
For a non-dividend-paying asset with continuously compounded risk-free rate \(r\), the risk-neutral model sets \(\gamma=r\)
On \(0=t_0<t_1<\cdots<t_d=T\), define \(\Delta t_j:=t_j-t_{j-1}\)
For \(Z_1,\ldots,Z_d\ \IIDsim\ \Norm(0,1)\),
\[ S_j=S_{j-1}\exp\!\left[ \left(r-\frac{\sigma^2}{2}\right)\Delta t_j +\sigma\sqrt{\Delta t_j}\,Z_j \right], \qquad j=1,\ldots,d \]
For equally spaced monitoring times, \(\Delta t_j=T/d\)
The simulated vector \((S_1,\ldots,S_d)\) is the discrete asset path used to evaluate option payoffs
Let \(\vX=(B(t_1),\ldots,B(t_d))\) be the target Brownian path
The function \(f\) builds the asset path and returns its discounted payoff
\[ Y=f(\vX)=\text{discounted payoff},\qquad \mu=\Ex(Y) \]
\[\begin{align*} K &= \text{the }\alert{\text{strike price}}, & T &= \text{the }\alert{\text{time to maturity}},\\ t_j &= jT/d, & S_j &= S(t_j), \qquad j=0,\ldots,d \end{align*}\]
| Contract | Discounted payoff |
|---|---|
| European call | \(\me^{-rT}\max\{S_d-K,0\}\) |
| European put | \(\me^{-rT}\max\{K-S_d,0\}\) |
| Arithmetic Asian call | \(\displaystyle \me^{-rT}\max\left\{\frac1T\int_0^T S(t)\,\dif t-K,0\right\}\approx \me^{-rT}\max\{A_d-K,0\}\) |
QMCPy FinancialOption offers the right and trapezoidal rules:
\[ \begin{aligned} A_d^{\mathrm{right}} &=\frac1d\sum_{j=1}^d S_j, & A_d^{\mathrm{trap}} &=\frac1d\left(\frac{S_0}{2}+\sum_{j=1}^{d-1}S_j+\frac{S_d}{2}\right) \end{aligned} \]
Asian, lookback, and barrier payoffs depend on the monitored path, not only the terminal price \(S_d\)
At the monitoring times define
\[ m_d:=\min_{0\le j\le d}S_j, \qquad M_d:=\max_{0\le j\le d}S_j \]
For a barrier \(B\), let \(\mathcal A_B\) be the event that the discretely monitored path activates the option
QMCPy takes \(S_0<B\) for an up barrier and \(S_0>B\) for a down barrier
\[ \begin{array}{ll} \text{up-in}: \mathcal A_B=\{M_d\ge B\}, & \text{up-out}: \mathcal A_B=\{M_d<B\},\\ \text{down-in}: \mathcal A_B=\{m_d\le B\}, & \text{down-out}: \mathcal A_B=\{m_d>B\} \end{array} \]
Using \(m_d\), \(M_d\), and \(\mathcal A_B\) from the preceding slide:
| Contract | Discounted payoff |
|---|---|
| Lookback call | \(\me^{-rT}(S_d-m_d)\) |
| Lookback put | \(\me^{-rT}(M_d-S_d)\) |
| Barrier call | \(\me^{-rT}\max\{S_d-K,0\}\indic(\mathcal A_B)\) |
| Barrier put | \(\me^{-rT}\max\{K-S_d,0\}\indic(\mathcal A_B)\) |
At the exercise times \(t_0,\ldots,t_d\), the holder chooses a stopping index \(J\) using only prices observed through \(t_J\)
Let \(\mathcal T_d\) be the set of these stopping indices. The fair price is
\[ \mu_{\mathrm{AmPut}} =\sup_{J\in\mathcal T_d} \Ex\!\left[ \me^{-rt_J}\max\{K-S_J,0\} \right] \]
Relaxing the requirement that \(\vU_0,\ldots,\vU_{n-1}\) be IID allows carefully constructed samples to cover \([0,1]^d\) more evenly
\[ \vU_0,\ldots,\vU_{n-1}\ \LDsim\ \Unif[0,1]^d \]
The empirical distribution of the sample is close to the uniform distribution
Discrepancy quantifies that difference; precise definitions come later
Low discrepancy replaces independent placement by deliberate coverage
Important families include
They are available in deterministic and randomized forms through SciPy and QMCPy
In software arrays rows correspond to samples and columns to coordinates
Return to the PCA factorization:
\[ \vX-\vmu =\sqrt{2+\sqrt2}\,\vv_1\Phi^{-1}(U_1) +\sqrt{2-\sqrt2}\,\vv_2\Phi^{-1}(U_2) \]
so \(U_1\) drives \(\lambda_1/(\lambda_1+\lambda_2)=(2+\sqrt2)/4\approx85.4\%\) of the total marginal variance
Put the largest eigenvalues first so dominant directions use the low-numbered coordinates, where many low discrepancy constructions are strongest
The quantile transform is a one-dimensional transport
In \(d\) dimensions
The map transports the proposal to the target when \(\varrho_{\mathrm{tar}}\bigl(\vT(\vz)\bigr) \left|\det\nabla \vT(\vz)\right| =\varrho_{\mathrm{prop}}(\vz)\)
Then, for every integrable function \(f\),
\[\begin{align*} \Ex_{\mathrm{tar}}[f(\vX)] &=\Ex_{\mathrm{prop}}[f\bigl(\vT(\vZ)\bigr)]\\ &\approx \frac1n\sum_{i=0}^{n-1}f\bigl(\vT(\vZ_i)\bigr), \qquad \vZ_0,\ldots,\vZ_{n-1}\ \IIDsim\ \varrho_{\mathrm{prop}} \end{align*}\]
An exact transport moves every proposal sample and makes the correction weight equal to one
Normalizing flow builds a flexible transport by composing manageable invertible maps, \(\vT=\vT_m\circ\cdots\circ \vT_1\)
The density equation supplies the constraint
The designer supplies the structure that makes the map solvable and useful
Let the proposal and target densities be
\[ \varrho_{\mathrm{prop}}(z)=1,\quad 0<z<1, \qquad \varrho_{\mathrm{tar}}(x)=2x,\quad 0<x<1 \]
The target is \(\operatorname{Beta}(2,1)\) with \(F_{\mathrm{tar}}(x)=x^2\), so its quantile transport is
\[ T(z)=F_{\mathrm{tar}}^{-1}(z)=\sqrt z, \qquad X=T(Z),\quad Z\sim\Unif(0,1) \]
Indeed, for \(0<z<1\),
\[ \varrho_{\mathrm{tar}}\bigl(T(z)\bigr)T'(z) =2\sqrt z\,\frac{1}{2\sqrt z} =1 =\varrho_{\mathrm{prop}}(z) \]
Thus, for every integrable \(f\),
\[ \Ex_{\mathrm{tar}}[f(X)] =\Ex_{\mathrm{prop}}\!\left[f\bigl(\sqrt Z\bigr)\right] \]
Let \(Z_1,Z_2\ \IIDsim\ \Norm(0,1)\) and, for \(b>0\), define
\[ X_1=Z_1, \qquad X_2=Z_2+b(Z_1^2-1) \]
This nonlinear map is easy to invert and has a simple Jacobian:
\[ \vT_b^{-1}(\vx) =\begin{pmatrix}x_1\\x_2-b(x_1^2-1)\end{pmatrix}, \qquad \nabla \vT_b(\vz) =\begin{pmatrix}1&0\\2bz_1&1\end{pmatrix}, \qquad \det\nabla \vT_b=1 \]
Thus, with \(\phi\) the standard normal density,
\[ \varrho_{\mathrm{tar}}(x_1,x_2) =\phi(x_1)\phi\!\left(x_2-b(x_1^2-1)\right), \qquad \Ex(X_2\mid X_1=x_1)=b(x_1^2-1) \]
Ingredients
Sampler for \(\vX_0,\ldots,\vX_{n-1}\)
\[ \begin{aligned} &i\leftarrow0\\[-0.2em] &\texttt{repeat}\\[-0.2em] &\qquad \texttt{repeat}\quad \vZ\sim\varrho_{\mathrm{prop}},\quad V\sim\Unif[0,1] \quad\text{independently}\\[-0.2em] &\qquad \texttt{until}\quad V\le\frac{\widetilde\varrho_{\mathrm{tar}}(\vZ)} {M\varrho_{\mathrm{prop}}(\vZ)}\\[-0.2em] &\qquad \vX_i\leftarrow \vZ,\quad i\leftarrow i+1\\[-0.2em] &\texttt{until}\quad i=n \end{aligned} \]
No normalizing constant \(c\) needed
Acceptance indicator
\[ I:= \begin{cases} 1, & V\le \widetilde\varrho_{\mathrm{tar}}(\vZ)/ \bigl[M\varrho_{\mathrm{prop}}(\vZ)\bigr],\\ 0, & \text{otherwise} \end{cases} \]
Density of an accepted proposal
\[\begin{align*} \varrho_{\mathrm{tar}}(\vz) &=\varrho_{\vZ\mid I}(\vz\mid1)\\ &=\frac{\Prob(I=1\mid \vZ=\vz)\varrho_{\mathrm{prop}}(\vz)} {\Prob(I=1)} &&\text{by Bayes' theorem}\\ &=\frac{\left(\widetilde\varrho_{\mathrm{tar}}(\vz)/ \bigl[M\varrho_{\mathrm{prop}}(\vz)\bigr]\right) \varrho_{\mathrm{prop}}(\vz)} {\Prob(I=1)} &&\text{by the acceptance rule}\\ &=\frac{\widetilde\varrho_{\mathrm{tar}}(\vz)}{M\Prob(I=1)} =c\widetilde\varrho_{\mathrm{tar}}(\vz) &&\text{since }c\widetilde\varrho_{\mathrm{tar}}\text{ is a density} \end{align*}\]
Uniform proposal and triangular target · Notebook
\[ \frac{\varrho_{\mathrm{tar}}(z)} {\varrho_{\mathrm{prop}}(z)} =2z\le2 \]
\[ V\le \frac{\varrho_{\mathrm{tar}}(Z)} {M\varrho_{\mathrm{prop}}(Z)} =Z \]
One uniform input, with the last coordinate making the decision
\[\begin{gather*} (\vU,V)=(U_1,\ldots,U_d,U_{d+1})\sim\Unif[0,1]^{d+1}, \\ \vZ=\vS(\vU)\sim\varrho_{\mathrm{prop}} \end{gather*}\]
For numerical integration: points \((\vu_i,v_i)\in[0,1]^{d+1}\), \(i=0,\ldots,n-1\)
Use IID uniform draws or randomized low discrepancy points
Target expectations become integrals over \([0,1]^{d+1}\)
Low discrepancy sampling starts with uniform points in this cube; performance depends on the resulting integrand, including its decision boundary
For \(0<p<1\), let \(\vT_k(\vZ)\sim\varrho_k\) and define
\[\begin{gather*} C(V)=\begin{cases}1,&V\le p,\\2,&V>p,\end{cases} \qquad \vX=\vT_{C(V)}\bigl(\vS(\vU)\bigr) \end{gather*}\]
For every integrable \(f\),
\[\begin{align*} \Ex_{\mathrm{mix}}[f(\vX)] &=\int_{[0,1]^{d+1}} f\!\left(\vT_{C(v)}\bigl(\vS(\vu)\bigr)\right) \,\dif\vu\,\dif v\\ &=\int_{[0,1]^d}\left[ \int_0^p f\bigl(\vT_1(\vS(\vu))\bigr)\,\dif v +\int_p^1 f\bigl(\vT_2(\vS(\vu))\bigr)\,\dif v\right]\dif\vu\\ &=p\Ex_{\varrho_1}[f(\vX)]+(1-p)\Ex_{\varrho_2}[f(\vX)]\\ &\approx\frac1n\sum_{i=0}^{n-1} f\!\left(\vT_{C(v_i)}\bigl(\vS(\vu_i)\bigr)\right) \end{align*}\]
The decision coordinate \(V\) selects a component at the fixed threshold \(p\)
Target \(\varrho_{\mathrm{tar}}=c\widetilde\varrho_{\mathrm{tar}}\), with \(\widetilde\varrho_{\mathrm{tar}}\le M\varrho_{\mathrm{prop}}\); define
\[ a(\vz):=\frac{\widetilde\varrho_{\mathrm{tar}}(\vz)} {M\varrho_{\mathrm{prop}}(\vz)}, \qquad I_i:=\boldsymbol{1}\!\left\{v_i\le a\bigl(\vS(\vu_i)\bigr)\right\} \]
\[\begin{align*} \Ex_{\mathrm{tar}}[f(\vX)] &=\frac{\displaystyle\int_{[0,1]^{d+1}} f\bigl(\vS(\vu)\bigr)\boldsymbol{1}\!\left\{v\le a\bigl(\vS(\vu)\bigr)\right\} \,\dif\vu\,\dif v} {\displaystyle\int_{[0,1]^{d+1}} \boldsymbol{1}\!\left\{v\le a\bigl(\vS(\vu)\bigr)\right\} \,\dif\vu\,\dif v}\\ &\approx\frac{\displaystyle\frac1n\sum_{i=0}^{n-1}f\bigl(\vS(\vu_i)\bigr)I_i} {\displaystyle\frac1n\sum_{i=0}^{n-1}I_i} =\frac{\displaystyle\sum_{i:\,I_i=1}f\bigl(\vS(\vu_i)\bigr)} {\displaystyle\sum_{i=0}^{n-1}I_i} \end{align*}\]
Two sample means, same points; their quotient averages accepted proposals
Defined when at least one proposal is accepted
Companion GeneratingSamples notebook
From travel-time uncertainty to option pricing: transform random inputs into outputs whose means, probabilities, or quantiles answer the application question
Direct methods turn uniform inputs into target samples, but a convenient transform or envelope may be hard to find
Markov Chain Monte Carlo takes a different route:
© 2026 Fred J. Hickernell · Illinois Tech · assisted by ChatGPT and Codex · Generating Samples · MATH 565 — Fall 2026 Website · \(\exstar\) = exercise