Expected value of a product of two correlated quantities

Gives the expected value of a product, such as the number of admissions times the cost per admission, from the two means and their covariance. The identity is exact for any joint distribution with finite variances. The product of the means equals the expected product only when the covariance is zero, for example when the two quantities are independent; a positive covariance raises the expected product above the product of the means and a negative one lowers it.

Signature

E_NU = E_N * E_U + Cov_NU
Inputs
InputsDefinitionUnit
E_NExpected number of admissions per patientadmissions per patient
E_UExpected cost per admissionpounds per admission
Cov_NUCovariance of the number of admissions and the cost per admission, the expected product of their deviations from their means, equal to their correlation times the two standard deviationspounds per patient
Output
E_NUExpected value of the product of the two quantities, here the expected admission cost per patientpounds per patient

Function

Expected value of an uncertain cost, health outcome or model output

Maps the probability distribution of an uncertain quantity, such as a cost per patient, a QALY total or a model output that depends on uncertain parameters, to its probability-weighted mean, which carries the units of the quantity. The discrete form applied at chance nodes is HE-FM-CHN-001 on the chance node page, the Monte Carlo mean over probabilistic simulations is HE-FM-ENB-001 and the constant-hazard event probability used below is HE-FM-TP-001. The records here cover the cases in which the mean of a function differs from the function of the means: a product of correlated quantities, a curved output with an uncertain parameter, an event probability under a gamma-distributed hazard and the mean of a log-normal cost. Notation follows the Expected Value article.

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Implementations

  • Excel

    Expected product from summary cells and from paired draws

    With MeanAdm, MeanUnitCost and CovAdmCost named, the first formula returns the expected product. With paired draws or patient records in the ranges AdmDraws and CostDraws, the second returns the mean of the products directly, which equals the first when the means and the population covariance COVARIANCE.P are taken from the same draws.

    =MeanAdm*MeanUnitCost+CovAdmCost; =SUMPRODUCT(AdmDraws,CostDraws)/COUNT(AdmDraws)

Assumptions

  • Means and covariance taken from one joint distribution

    E_N, E_U and Cov_NU describe the same population or the same set of paired simulation draws. Taking the two means from different sources and setting the covariance to zero assumes independence rather than establishing it.

  • Finite variances for the product identity

    Both quantities have finite variances, so the covariance exists. No distributional form and no independence is needed; with independence the identity reduces to the product of the means.

Worked examples

  • Admissions and cost per admission with a covariance of 300

    With a mean of 1.5 admissions, a mean of £3,000 per admission and a covariance of 300, the expected cost is 1.5 x 3,000 + 300 = 4,800 pounds per patient, against £4,500 from multiplying the means, as in the article. The figures are illustrative.

    E_N = 1.5; E_U = 3000; Cov_NU = 300; E_NU = 4800
  • Same admission means with independent counts and unit costs

    When admissions and cost per admission are independent the covariance is zero and the expected cost is the product of the means, £4,500 per patient, so the covariance of 300 accounts for the whole gap of £300 in the first example.

    E_N = 1.5; E_U = 3000; Cov_NU = 0; E_NU = 4500
  • Same admission means with a negative covariance of 300

    If patients with more admissions had cheaper admissions, a covariance of minus 300 would lower the expected cost to £4,200 per patient, below the product of the means; correlation can narrow or widen the gap, as the article notes. The figure is computed here for illustration.

    E_N = 1.5; E_U = 3000; Cov_NU = -300; E_NU = 4200

Common errors

  • Multiplying mean admissions by mean cost per admission

    Multiplying the two means drops the covariance: £4,500 in place of £4,800 per patient in the article's example, an understatement of 6.25 per cent of the expected cost that carries into every budget built on it.

  • Combining admission counts and unit costs from separate sources as if independent

    When counts and unit costs come from different studies, the covariance cannot be estimated and is set to zero without being stated. The resulting expected cost rests on an independence assumption that patient-level data recording both quantities could check.

Sources

  • Covariance identity behind the expected admission cost

    Casella G, Berger RL. Statistical Inference. 2nd ed. Pacific Grove, CA: Duxbury; 2002 (reprinted Boca Raton: Chapman and Hall/CRC; 2024). Section 4.5, covariance and correlation: the covariance of X and Y equals E[XY] minus E[X]E[Y], so the expected product is the product of the means plus the covariance; independent variables have zero covariance. Textbook result.

    View source →

Canonical Identity

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