Assessing the Impact of Osteoporosis on the Burden of Hip Fractures

Abstract

The aim of the study was to determine the number of hip fractures within defined countries for 2010 and the proportion attributable to osteoporosis. The number of incident hip fractures in one year in countries for which data were available was calculated from the population demography in 2010 and the age- and sex-specific risk of hip fracture. The number of hip fractures attributed to osteoporosis was computed as the number of hip fractures that would be saved assuming that no individual could have a femoral neck T-score of less than −2.5 SD (i.e., the lowest attainable T-score was that at the threshold of osteoporosis (=−2.5 SD). The total number of new hip fractures for 58 countries was 2.32 million (741,005 in men and 1,578,809 in women) with a female-to-male ratio of 2.13. Of these 1,159,727 (50 %) would be saved if bone mineral density in individuals with osteoporosis were set at a T-score of −2.5 SD. The majority (83 %) of these “prevented” hip fractures were found in men and women at the age of 70 years or more. The 58 countries assessed accounted for 83.5 % of the world population aged 50 years or more. Extrapolation to the world population using age- and sex-specific rates gave an estimated number of hip fractures of approximately 2.7 million in 2010, of which 1,364,717 were preventable with the avoidance of osteoporosis (264,162 in men and 1,100,555 in women). We conclude that osteoporosis accounts for approximately half of all hip fractures. Strategies to prevent osteoporosis could save up to 50 % of all hip fractures.

Appendix

The Hazard Function of Fracture According to BMD at the Femoral Neck

We assume that BMD has a distribution, which is very close to normal. Let h denote the hazard function of hip fracture of a randomly chosen individual from the population at a certain age and sex, and let σ be the standard deviation of the random variable BMD. The GR describes the increase in hip fracture risk for each SD decrease in femoral neck BMD. Then, if BMD is also specified and equal to z, the hazard function is

$$ {\text{h}} \cdot { \exp }(( - { \log }\left( {\text{GR}} \right)/{{\upsigma}}) \cdot {\text{z}})/{\text{E}}[{ \exp }(( - { \log }\left( {\text{GR}} \right)/\sigma ) \cdot {\text{BMD}})] $$

(1)

where the denominator E[exp ((−log (GR)/σ)·BMD)] is the expected value of exp((−log (GR)/σ)·BMD). The denominator is needed to make the expected value of equation (1) (the mean risk) equal to h. We can derive from equation (1) the hazard ratio when comparing the BMD value z − σ and z, which differ exactly 1 SD, is exp (log (GR)) = GR. Now we have to determine E[exp ((−log (GR)/σ)·BMD)]. For a random variable Y, which has a normal distribution with mean m and standard deviation SD, the following relationship is true:

$$ E\left[ {\exp \, \left( Y \right)} \right] \, = \, \exp \left( {m \, + {\text{ SD}}^{2} /2} \right) $$

(2)

In order to realize that E[exp (Y)] exp (m), we can apply Jensen’s inequality, which states that E[g(V)] > g(E[V]) for any random variable V when g is a convex function.

By applying relationship (2), we find

$$ E[\exp \, (( - \log \left( {\text{GR}} \right)/\sigma ) \times {\text{BMD}})] \, = \, \exp \;\,(( - \log \left( {\text{GR}} \right)/\sigma ) \times E\left[ {\text{BMD}} \right] + \left( {\log \left( {\text{GR}} \right)} \right)^{2} /2) $$

Thus,

$$ 1/ \, E[\exp \,(( - \log \,\;\left( {\text{GR}} \right)/\sigma ) \cdot {\text{BMD)}}] \, = \, \exp \,((\log \,\left( {GR} \right)/\sigma ) \cdot E\left[ {\text{BMD}} \right] - \left( {\log \,\left( {\text{GR}} \right)} \right)^{2} /2) $$

and expression (1) equals

$$ h \cdot { \exp }( - { \log }\left( {\text{GR}} \right) \cdot \left( {z - E\left[ {\text{BMD}} \right]} \right)/\sigma -- \, \left( {{ \log }\left( {\text{GR}} \right)} \right)^{ 2} / 2) $$

If the linear transformation t(z) = (z − E[BMD])/σ of z is used, then equation (1) can be written as:

$$ h \cdot { \exp }\,( - { \log }\,\left( {\text{GR}} \right)\, \times \,t \, -- \, \left( {{ \log }\,\,\left( {\text{GR}} \right)} \right)^{ 2} / 2) $$

If all individuals below a limit g of BMD would be carried to the BMD value equal to g and the other individuals are unchanged, then how would the risk change? First we can note that t(g) = (g − E[BMD])/σ. The calculated risk is

$$ h \cdot \exp \,( - \log (GR) \cdot t(g) - (\log ({\text{GR}}))^{2} /2) \cdot \int\limits_{ - \infty }^{t(g)} {\frac{1}{{\sqrt {2 \cdot \pi } }} \cdot \exp ( - t^{2} /2){\text{d}}t + } $$

$$ h \cdot \int\limits_{t(g)}^{\infty } {\frac{1}{{\sqrt {2 \cdot \pi } }} \cdot \exp \,( - \log ({\text{GR}}) \cdot t - (\log } (GR))^{2} /2) \cdot \exp \,( - t^{2} /2)dt $$

The second term equals

$$ h \cdot \int\limits_{t(g)}^{\infty } {\frac{1}{{\sqrt {2 \cdot \pi } }} \cdot \exp \,( - \frac{1}{2} \cdot (t + \log \,(GR))^{2} ){\text{d}}t} $$

That implies that the risk (the hazard function) is

$$ {\text{h}} \cdot \{ { \exp }( - { \log }\,\left( {\text{GR}} \right) \cdot t\left( g \right) - \left( {{ \log }\,\left( {\text{GR}} \right)} \right)^{ 2} / 2) \cdot \Upphi \left( {t\left( g \right)} \right) \, + { 1 } - \Upphi \left( {t\left( g \right) + { \log }\,\left( {\text{GR}} \right)} \right)\} $$

(3)

where Φ is the standardized normal distribution function.

The factor with which h is multiplied in equation (3) was calculated for the different gradients given in Johnell et al. [21] and the limit g = 0.558 g/cm².

We note that

$$ t\left( g \right) \, = \, \left( {g - E\left[ {\text{BMD}} \right]} \right)/\sigma $$

The mean value E[BMD] and the standard deviation σ vary by age and sex. The limit 0.558 fulfils Φ(t(0.558)) = 0.0062, when the mean and standard deviation equals that of women in the age interval 20–29 years. The limit corresponds to 2.5 SD below the mean of women 20–29 years of age.