Why Sensitivity Matters
A model gives a prediction from chosen inputs, such as an initial stock and parameters describing a rate. Those inputs are rarely known with perfect precision. Sensitivity analysis asks how much the prediction changes when one input changes, while the others are held fixed.
This is more informative than simply reporting a predicted total. If a small change in one parameter produces a large change in the prediction, uncertainty in that parameter deserves attention. If changing the starting stock shifts the result by a fixed amount but changing a rate parameter changes the accumulated total substantially, the model identifies different sources of sensitivity.
We will study a fixed-time accumulation model. The rate may depend on time and on a parameter, but in this model it does not depend on the current stock. That assumption is important: it makes the effect of the initial stock especially direct.
A Model for Accumulated Totals
Let \(S_0\) be the starting stock, let \(T\) be the prediction time, and let \(r(t;\theta)\) be the rate at time \(t\), with \(\theta\) a rate parameter. The predicted stock at time \(T\) is the initial stock plus the accumulated rate:
Here \(T\) is fixed when we compare input values. The integral is the net amount added over the interval; a negative rate would subtract from the stock. The formula assumes that the rate is integrable on \([0,T]\). If the rate is nonnegative, the accumulated amount is nonnegative as well.
A sensitivity is a local rate of change, not necessarily the exact effect of a finite change. For a small change \(\Delta\theta\), the derivative gives the approximation \(\Delta S\approx \frac{\partial S}{\partial\theta}\Delta\theta\), provided the model is differentiable near the chosen parameter value. For a larger change, evaluate the model at both parameter values rather than relying only on this approximation.
Why the Sensitivity Formula Works
For the model above, the initial stock enters as a separate additive term. The rate parameter enters inside the accumulated rate. If \(r(t;\theta)\) and its partial derivative with respect to \(\theta\) are continuous on the relevant time and parameter ranges, differentiation under the integral sign gives the following result.
The first formula follows directly: increasing \(S_0\) by one stock unit adds one stock unit to the prediction, because it does not change the rate in this model. For the second formula, a small parameter change changes the rate at each time, and those rate changes accumulate over the full interval.
Compare \(S(T;S_0,\theta+h)\) and \(S(T;S_0,\theta)\). The initial-stock terms cancel, leaving the difference of the two rate integrals.
Divide that difference by \(h\). The quotient is the integral of \(\frac{r(t;\theta+h)-r(t;\theta)}{h}\).
As \(h\) tends to zero, the integrand tends to \(\frac{\partial r}{\partial\theta}(t;\theta)\). Continuity on the fixed interval permits passing this limit through the integral, giving the stated formula.
This result identifies where rate-parameter sensitivity comes from: the model's predicted change is the integral of the rate's own sensitivity over time. Positive and negative contributions can reinforce or offset one another.
Worked Example: Sensitivity to Starting Stock and a Rate Multiplier
Worked Example: Sensitivity to Starting Stock and a Rate Multiplier
Suppose the rate is \(r(t;k)=k(1+t)\), where \(t\) is numerical time in days, \(k\) has units of stock per day, and \(S_0=10\) stock units. Find the predicted stock after \(T=3\) days when \(k=2\), then find the sensitivities to \(S_0\) and \(k\).
First integrate the rate. Since \(\int_0^3(1+t)\,dt=[t+t^2/2]_0^3=3+9/2=10.5\) days, the model is
At the specified values, the accumulated amount is \(10.5(2)=21\) stock units, so \(S(3;10,2)=10+21=31\) stock units.
The starting-stock sensitivity is \(\frac{\partial S}{\partial S_0}=1\). The rate-parameter sensitivity is \(\frac{\partial S}{\partial k}=10.5\) days. Thus, near \(k=2\), an increase of \(0.1\) in \(k\) predicts an increase of approximately \(10.5(0.1)=1.05\) stock units.
In this example the approximation is exact, because the model is linear in \(k\). Substitution confirms it: \(S(3;10,2.1)=10+10.5(2.1)=10+22.05=32.05\), which is \(32.05-31=1.05\) stock units above the original prediction.
When a Larger Rate Parameter Lowers the Total
A parameter's effect depends on how it enters the rate law. A larger parameter does not necessarily mean a larger accumulated total. If the parameter makes the rate decline more quickly, its sensitivity can be negative.
Worked Example: Sensitivity to a Rate-Decay Parameter
Consider \(r(t;k)=a(1-kt)\) on \(0\leq t\leq T\), where \(a=6\) stock units per day and \(k=0.2\) per day. Let \(S_0=4\) stock units and \(T=2\) days. The rate remains nonnegative throughout this interval because \(1-kT=1-0.4=0.6\geq0\).
Integrating gives
At the stated values, the accumulated amount is \(6(2-0.2(4)/2)=6(2-0.4)=6(1.6)=9.6\) stock units. The predicted total is therefore \(4+9.6=13.6\) stock units.
Differentiate with respect to \(k\): \(\frac{\partial S}{\partial k}=-aT^2/2\). At \(a=6\) and \(T=2\), this equals \(-6(4)/2=-12\) stock units per unit of \(k\). The negative sign means that increasing \(k\) lowers the predicted total in this model.
For a finite check, increase \(k\) by \(0.05\), from \(0.20\) to \(0.25\) per day. The derivative predicts a change of \((-12)(0.05)=-0.6\) stock units. Direct substitution gives \(S(2;4,0.25)=4+6(2-0.25(4)/2)=4+6(1.5)=13\) stock units, a decrease of \(13-13.6=-0.6\) stock units. The result matches the prediction exactly because this model is linear in \(k\).
Comparing Percentage Changes With Elasticity
Absolute sensitivities have units, so their sizes cannot always be compared directly across inputs with different units or scales. Elasticity provides a dimensionless comparison. For a positive input \(x\) and a positive predicted total \(S\), define the elasticity of \(S\) with respect to \(x\) by \(E_x=\frac{x}{S}\frac{\partial S}{\partial x}\). Locally, a 1% increase in \(x\) produces approximately an \(E_x\)% change in \(S\). This is a local approximation, not a guarantee for a large change.
Worked Example: Comparing Relative Sensitivity
Suppose \(r(t;k)=k^2(1+t)\), \(S_0=5\), \(T=2\), and \(k=3\). Time \(t\) is measured numerically in days, and the units of \(k^2\) make the rate a stock amount per day.
The integral factor is \(\int_0^2(1+t)\,dt=2+2=4\). Therefore \(S(2;S_0,k)=S_0+4k^2\). At the given inputs, \(S=5+4(9)=41\) stock units.
The partial derivatives are \(\frac{\partial S}{\partial S_0}=1\) and \(\frac{\partial S}{\partial k}=8k\). At \(k=3\), the latter is \(24\) stock units per unit of \(k\). The elasticities are consequently \(E_{S_0}=(5/41)(1)=5/41\approx0.122\) and \(E_k=(3/41)(24)=72/41\approx1.756\). Thus, locally, a 1% increase in starting stock corresponds to about a 0.122% increase in the prediction, whereas a 1% increase in \(k\) corresponds to about a 1.756% increase.
Check with a 1% increase in each input. Increasing \(k\) to \(3.03\) gives \(S=5+4(3.03)^2=5+4(9.1809)=41.7236\), an increase of \(0.7236/41\approx1.765\%\), close to the local estimate of 1.756%. Increasing \(S_0\) to \(5.05\) gives \(S=5.05+36=41.05\), an increase of \(0.05/41\approx0.122\%\), matching its elasticity to the displayed precision.
A Common Mistake: Ignoring the Rate's Parameter Dependence
A frequent error is to differentiate the initial stock correctly but treat the accumulated rate as if it did not depend on the parameter. For the first example, one might write \(S=S_0+10.5k\) and then claim \(\frac{\partial S}{\partial k}=0\), reasoning that only the initial stock is fixed. This is wrong: \(10.5k\) is part of the prediction and changes with \(k\).
The correction is to differentiate every term that depends on the input being varied. Here \(\frac{\partial}{\partial k}(S_0+10.5k)=0+10.5=10.5\), while the initial-stock derivative is \(\frac{\partial}{\partial S_0}(S_0+10.5k)=1+0=1\). Holding an input fixed means not changing it during a particular comparison; it does not mean discarding other terms from the derivative.
What the Sensitivity Does—and Does Not—Say
A derivative describes behavior near the chosen input values. It can reveal whether a prediction increases or decreases locally and how rapidly it changes there. For a finite input change, the exact difference is the difference between two model evaluations; a derivative-based estimate may be inaccurate if the model is strongly nonlinear over that change.
The result \(\frac{\partial S}{\partial S_0}=1\) has a specific condition: the rate in this model does not depend on the current stock or on \(S_0\). If changing the starting stock also changes the later rate, then that indirect effect must be included in the model before evaluating sensitivity. Likewise, a rate-parameter derivative depends on the chosen rate law and on the prediction time \(T\); it is not a universal property of the parameter's name.
Checking units and signs is a useful safeguard. A sensitivity to a rate parameter has units of predicted stock divided by parameter units. A positive derivative indicates an increasing prediction locally; a negative derivative indicates a decreasing prediction locally. Neither sign alone establishes what happens for all possible parameter values.
Check Your Understanding
Use the model definitions and assumptions stated in each question. Keep track of which input is changing and which are held fixed.
- For \(S(T;S_0,\theta)=S_0+\int_0^T r(t;\theta)\,dt\), what assumption makes \(\frac{\partial S}{\partial S_0}=1\)?
- If \(\frac{\partial r}{\partial\theta}\) is positive throughout the integration interval, what can you conclude about \(\frac{\partial S}{\partial\theta}\), assuming the interval has positive length?
- For \(S(T;S_0,k)=S_0+a(T-kT^2/2)\), calculate \(\frac{\partial S}{\partial k}\) and explain its sign when \(a>0\) and \(T>0\).
- What information does elasticity provide that an absolute sensitivity may not provide as directly?
- Why should a finite parameter change be checked by evaluating the model at both parameter values rather than relying only on a derivative?
