The use of multiple sites for the diagnosis of osteoporosis

Abstract

Introduction

It has been suggested that bone mineral density (BMD) measurements should be made at multiple sites, and that the lowest T–score should be taken for the purpose of diagnosing osteoporosis.

Purpose

The aim of this study was to examine the use of BMD measurements at the femoral neck and lumbar spine alone and in combination for fracture prediction.

Methods

We studied 19,071 individuals (68% women) from six prospective population-based cohorts in whom BMD was measured at both sites and fracture outcomes documented over 73,499 patient years. BMD values were converted to Z-scores, and the gradient of risk for any osteoporotic fracture and for hip fracture was examined by using a Poisson model in each cohort and each gender separately. Results of the different studies were merged using weighted β-coefficients.

Results

The gradients of risk for osteoporotic fracture and for hip fracture were similar in men and women. In men and women combined, the risk of any osteoporotic fracture increased by 1.51 [95% confidence interval (CI)=1.42–1.61] per standard deviation (SD) decrease in femoral-neck BMD. For measurements made at the lumbar spine, the gradient of risk was 1.47 (95% CI=1.38–1.56). Where the minimum of the two values was used, the gradient of risk was similar (1.55; 95% CI=1.45–1.64). Higher gradients of risk were observed for hip fracture outcomes: with BMD at the femoral neck, the gradient of risk was 2.45 (95% CI=2.10–2.87), with lumbar BMD was 1.57 (95% CI=1.36–1.82), and with the minimum value of either femoral neck and lumbar spine was 2.11 (95% CI=1.81–2.45). Thus, selecting the lowest value for BMD at either the femoral neck or lumbar spine did not increase the predictive ability of BMD tests. By contrast, the sensitivity increased so that more individuals were identified but at the expense of specificity. Thus, the same effect could be achieved by using a less stringent T–score for the diagnosis of osteoporosis.

Conclusions

Since taking the minimum value of the two measurements does not improve predictive ability, its clinical utility for the diagnosis of osteoporosis is low.

Appendix

The proportion of individuals with at least one type of BMD measurements below a cut off

Let A_i denote the event that the measurement of type i is below a cut off g. We use the symbols ∪ and ∩ for the union and the intersection of events, respectively. The probability that any of n measurements is below the cut off is

$${\text{P}}{\left( {{\text{A}}_{1} \, \cup \,{\text{A}}_{2} \, \cup \, \ldots \, \cup \,{\text{A}}_{n} } \right)}$$

The following notations are used

$$\begin{array}{*{20}c} {{\text{S}}_{{\text{1}}} {\text{ = }}{\sum {{\text{P}}{\left( {{\text{A}}_{{\text{i}}} } \right)}} }} \\ {{\text{S}}_{{\text{2}}} {\text{ = }}{\sum {{\text{P}}{\left( {{\text{A}}_{{\text{i}}} \cap {\text{A}}_{{\text{j}}} } \right)}} }} \\ {{\text{S}}_{{\text{3}}} {\text{ = }}{\sum {{\text{P}}{\left( {{\text{A}}_{{\text{i}}} \cup {\text{A}}_{{\text{j}}} \cap {\text{A}}_{{\text{k}}} } \right)}} }} \\ {{\text{S}}_{{\text{n}}} {\text{ = }}{\sum {{\text{P}}{\left( {{\text{A}}_{{\text{1}}} \cap {\text{A}}_{{\text{2}}} {\text{ \ldots }} \cap {\text{A}}_{{\text{n}}} } \right)}} }} \\ \end{array} $$

Then the relationship below holds

$${\text{P}}{\left( {{\text{A}}_{{\text{1}}} \, \cup {\text{A}}_{{\text{2}}} \, \cup {\text{ \ldots }}\, \cup \,{\text{A}}_{{\text{n}}} } \right)}{\text{ = }}{\sum {{\left( {{\text{ - 1}}} \right)}^{{{\text{i + 1}}}} \,{\text{S}}_{{\text{i}}} } }$$

The argument (−1)ⁱ⁺¹ obviously can only take the values +1 or −1. This is because the first argument of the summation (S₁) double counts some combination areas of A₁ A₂ A₃ etc. Those probabilities have to be subtracted in the second argument, but since too much is subtracted, there needs to be some adding in the third argument and so on…

When only two sites of measurement are considered (e.g. BMD at lumbar spine and femoral neck), the relationship can be written

$${\text{P}}{\left( {{\text{A}}\, \cup \,{\text{B}}} \right)}{\text{ = P}}{\left( {\text{A}} \right)}{\text{ + P}}{\left( {\text{B}} \right)}{\text{ - P}}{\left( {{\text{A}} \cap {\text{B}}} \right)}$$

The last term of the above expression, P(A∩B), is the probability that both measurements are below the cut-off limit g. We assume that the scale of the measurements is standardised so their means and SDs are 0 and 1, respectively. If the correlation coefficient between the measurements is ρ, then the probability that both of two measurements, X and Z, are below the limit g and can be written

$${\text{P}}{\left( {{\text{A}} \cap {\text{B}}} \right)} = {\text{P}}{\left( {{\left\{ {{\text{X}} < {\text{g}}} \right\}} \cap {\left\{ {{\text{Z}} < {\text{g}}} \right\}}} \right)} = {\text{P}}{\left( {{\left\{ {{\text{X}} < {\text{g}}} \right\}} \cap {\left\{ {\rho \cdot {\text{X}} + {\text{Y}} < {\text{g}}} \right\}}} \right)}$$

, where X and Y are independent and X has a normal distribution with the mean 0 and the standard deviation 1, and Y has a normal distribution with the mean 0 and the variance 1−ρ ². Then the variance of ρ·X+Y is 1 and the covariance between X and ρ·X+Y is ρ, so the correlation coefficient is also ρ. Let Φ denote the standardised normal distribution function. We then obtain

$${\text{P}}{\left( {{\text{A}} \cap {\text{B}}} \right)} = {\int\limits_{ - \infty }^g {\Phi (\frac{{g - \rho \cdot {\text{x}}}}{{{\sqrt {1 - \rho ^{2} } }}}) \cdot \frac{{\exp ({\text{ - x}}^{{\text{2}}} {\text{/2}})}}{{{\sqrt {2\pi } }}}{\text{dx}}} }$$

and

$${\text{P}}{\left( {\text{A}} \right)} + {\text{P}}{\left( {\text{B}} \right)} = 2\,\,\Phi {\left( {\text{g}} \right)}$$

Thus, the probability that any of the two measurements is below the limit g, P(A ∪ B) can be calculated by the two last-mentioned relationships.