Why Dependence Matters in Decision Analysis
Probabilistic sensitivity analysis propagates uncertain inputs through a model to estimate quantities such as the probability of meeting a development target. The result depends not only on the uncertainty assigned to each input separately, but also on how the inputs vary together. If two inputs tend to be high at the same time, or one tends to be high when the other is low, independent draws can create combinations that are too common or too rare.
For example, an efficacy-related input and a safety-related input might both depend on a shared uncertain biological quantity. Sampling each from its own marginal distribution but independently would discard that shared relationship. Depending on how the decision rule combines the inputs, the estimated probability of success could then be too high or too low. The direction cannot be inferred from the word “dependence” alone; it depends on the decision function and the structure of the joint distribution.
The key task is therefore to represent and sample from a defensible joint distribution. This tutorial focuses on preserving that dependence when estimating decision probabilities. It does not prescribe which marginal distribution should be selected for every input; that choice is a separate modelling decision.
Covariance and a Joint Normal Model
For two inputs with finite variances, covariance records how their deviations from their means vary together: \( \operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])] \). Positive covariance indicates that above-mean values tend to occur together; negative covariance indicates that an above-mean value of one tends to accompany a below-mean value of the other. The correlation \( \rho \) is covariance divided by the product of the standard deviations, so it is dimensionless and lies between \(-1\) and \(1\).
A bivariate normal model is one useful joint model when a normal approximation is defensible. If \(X\) and \(Y\) each have mean zero and variance one, with correlation \(\rho\), then their covariance is also \(\rho\). For a linear combination, the covariance term directly determines its variance:
These identities follow by expanding the squared deviation. For the sum, write the deviation as \((X-E[X])+(Y-E[Y])\), square it, and take expectations. The two squared terms give the two variances; the cross-product contributes \(2E[(X-E[X])(Y-E[Y])]\), which is twice the covariance. For the difference, the cross-product has a minus sign, giving the second identity.
For standard normal inputs, a convenient way to generate draws with a specified correlation is to take independent standard normal draws \(Z_1,Z_2\) and set \(X=Z_1\) and \(Y=\rho Z_1+\sqrt{1-\rho^2}Z_2\). This produces standard normal marginals and covariance \(\rho\). The construction requires \(|\rho|\leq1\); in higher dimensions, the proposed covariance matrix must be positive semidefinite to represent a valid joint distribution.
Worked Example: A Decision Depending on the Sum
Hypothetical illustration. Let \(X\) and \(Y\) be standardized uncertain contributions to a development decision. Suppose their joint distribution is bivariate normal, both have mean zero and variance one, and their correlation is \(0.75\). A decision criterion is met when \(X+Y>1.5\). Compare the probability under the joint model with the probability obtained by incorrectly drawing the same marginals independently.
Under the joint model, the variance of the sum is \(1+1+2(0.75)=3.5\). A linear combination of jointly normal variables is normal, so \(X+Y\sim N(0,3.5)\), where the second parameter is the variance. Therefore:
If the inputs are instead drawn independently, their variances remain one but their covariance is set to zero. The sum then has variance \(2\), giving:
For this hypothetical threshold, the independent model gives a probability about \(0.06692\) lower than the joint model: approximately \(21.13\%\) versus \(14.44\%\). The difference arises because positive dependence increases the variability of the sum. These are model-based probabilities under the stated distributions, not frequencies that apply automatically to a real development program.
Dependence Can Change a Decision in the Other Direction
The previous result is not a general rule that ignoring positive dependence lowers a decision probability. For a difference, positive covariance reduces the variance. If the decision depends on a large difference, independent draws can instead make the event appear more likely. The next calculation uses a distinct hypothetical correlation to make this direction clear.
Worked Example: A Decision Depending on a Difference
Hypothetical illustration. Suppose standardized inputs \(X\) and \(Y\) are jointly normal, each with mean zero and variance one, and correlation \(0.60\). The decision criterion is \(X-Y>1.5\). Compare the joint-model probability with the result from independent draws.
The difference has mean zero and variance \(1+1-2(0.60)=0.80\). Thus:
With independent draws, the variance of \(X-Y\) is \(1+1=2\), so:
Here independent draws overstate the probability by about \(0.09766\), or \(9.77\) percentage points. The arithmetic follows the variance identity: positive covariance subtracts from the variance of a difference. The example shows why the decision function must be considered alongside the dependence structure.
In practical MIDD analyses, inputs need not be normal, and the decision rule need not be a linear threshold. The general procedure remains the same: specify a joint model for the uncertain inputs, draw complete input vectors from it, evaluate the decision for each vector, and summarize the fraction satisfying the criterion. If \(M\) draws are used and \(I_m\) equals one when draw \(m\) meets the criterion and zero otherwise, the Monte Carlo estimate is:
This estimates the probability under the specified joint model. Increasing \(M\) reduces simulation error, but it cannot fix a joint model that misrepresents the dependence.
Representing and Sampling the Joint Structure
A joint distribution can be represented directly, conditionally, or through a dependence construction. For a small set of discrete inputs, a joint probability table may be transparent. A conditional model specifies one input’s distribution given another. For multivariate normal inputs, a mean vector and covariance matrix define the joint normal distribution. A Cholesky factorization can be used to transform independent standard normal draws into correlated draws, provided the covariance matrix is valid.
When marginals are specified separately and a normal dependence construction is used, a Gaussian copula is one option: correlated normal draws are transformed through the standard normal cumulative distribution function to uniform draws, then through the inverse cumulative distributions of the desired marginals. This retains the selected marginal distributions while imposing a dependence structure associated with the correlated normal variables. It does not guarantee that the resulting dependence matches all scientifically relevant features, such as joint tail behaviour.
Worked Example: Sampling a Joint Discrete Input Pair
Hypothetical illustration. Let \(A\) indicate whether one uncertain condition is present and \(B\) indicate whether a second condition is present. Suppose each is a binary input, and the proposed joint distribution is:
| \(A\) | \(B\) | Joint probability |
|---|---|---|
| 0 | 0 | 0.40 |
| 0 | 1 | 0.10 |
| 1 | 0 | 0.10 |
| 1 | 1 | 0.40 |
The probabilities sum to \(1.00\). Each marginal probability of a value of one is \(0.10+0.40=0.50\). If the decision criterion requires both conditions, its probability under the joint model is the joint cell probability \(P(A=1,B=1)=0.40\). Under independent draws with these same marginals, it would be \(0.50\times0.50=0.25\).
The association can also be checked numerically. Since \(A\) and \(B\) are binary, \(E[A]=E[B]=0.50\), \(E[AB]=0.40\), and hence \(\operatorname{Cov}(A,B)=0.40-(0.50)(0.50)=0.15\). Each variance is \(0.50(1-0.50)=0.25\), so the correlation is \(0.15/\sqrt{0.25\times0.25}=0.60\). To sample correctly, draw one of the four paired outcomes using its listed probability; do not sample \(A\) and \(B\) separately as independent Bernoulli variables. The example shows that a joint table can preserve dependence directly and transparently.
Diagnostics, Alternatives, and Limitations
A dependence specification is an assumption that should be checked against its scientific basis and the available evidence. Useful diagnostics include the following:
- Check the marginal draws. Confirm that each simulated input has the intended distribution, range, units, and summary characteristics.
- Check the joint draws. Plot paired draws or tabulate conditional patterns. Compare simulated correlation or other dependence summaries with the values intended by the model.
- Check constraints. Verify that combinations obey known structural restrictions. For example, paired probabilities must be nonnegative and sum to one; a covariance matrix must be positive semidefinite.
- Check the decision consequence. Recalculate the decision probability under plausible alternative dependence structures when dependence is poorly known. Label such results as sensitivity analyses, not as equally supported estimates unless the evidence warrants that interpretation.
A common failure is to verify only the marginal distributions. A simulated \(X\) histogram and \(Y\) histogram can each look correct even when the pairing is wrong. Another is to specify pairwise correlations that cannot jointly form a valid covariance matrix. Correlation itself is also incomplete: it measures linear association, and zero correlation does not in general imply independence. Two distributions can share the same marginals and correlation yet have different tail dependence and therefore different probabilities for an important decision event.
The main alternatives have different uses. Independent sampling is appropriate when inputs are genuinely independent under the uncertainty model, or when independence is deliberately used as a clearly labelled comparison. A joint distribution or conditional model is preferable when scientific knowledge supports particular combinations or conditional relationships. Correlated sampling is convenient for continuous inputs when a defensible covariance structure is available. A copula can keep chosen marginals while specifying a dependence construction, but adds modelling choices and may not represent tail relationships well.
No sampling technique can create information about dependence that is absent from the evidence. If dependence is uncertain, report that limitation and examine how reasonable joint structures affect the decision. Do not interpret an estimated correlation as a causal relationship, and do not describe an independence-based result as neutral: independence is itself a substantive joint assumption.
Common Mistake
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For standard normal inputs with correlation \(0.40\), what is the variance of their sum and the variance of their difference?
- Two binary inputs each have marginal probability \(0.30\) of being one. Under independence, what is the probability that both are one?
- Why can correct simulated marginal distributions fail to preserve the probability of a decision that depends on two inputs jointly?
- What validity condition must a covariance matrix meet before it can define a multivariate normal joint model?
Solutions. (1) The sum has variance \(1+1+2(0.40)=2.80\); the difference has variance \(1+1-2(0.40)=1.20\). (2) Independence gives \(0.30\times0.30=0.090\). (3) The marginal checks do not verify how the inputs are paired; the event probability can change with dependence even when the two marginal distributions are unchanged. (4) The covariance matrix must be positive semidefinite, so every linear combination has nonnegative variance.