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I am reading a text on survival analysis (Smith's 2002 Analysis of Failure and Survival Data). All concepts like hazard function, survival function, density of survival variable $Y$ are rigorously defined. However, in an exercise (p.14) it says "Suppose that the death rate of a person that smokes is at each age twice that of a non-smoker. If $h_s(y)$ denotes the hazard rate of a smoker at age $y$ and $h_n(y)$ that of a non-smoker at age $y$, write an equation relating $h_s(y)$ and $h_n(y)$".

This left me puzzled on what the author means by death rate (it has not been defined as far as I can see). Perhaps he means the derivative of the survival function or something alike.

This idea may be backed up by a google search in which I found the definition

$$\lambda = \frac{D}{T}$$

where $D$ number of deaths in time interval $T$. In probability terms I would thus conclude a useful definition may be

$$\lambda = \frac{F(Y+t) - F(Y) }{t}$$

where $F(Y)$ the cummulative distribution function. Furthermore if I let

$$\lim_{t \rightarrow 0} \frac{F(Y+t) - F(Y) }{t} = f(y)=-S'(Y)$$

where $f$ the density and $S$ the survival function.

Is this logic formalized or convention somewhere?

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A rate has a specific definition of $\frac{\# \mbox{events}}{\# \mbox{person-years}}$. A risk on the other hand refers to a particular individual's risk of experiencing an outcome of interest, and it is risk which is intrinsically related to the hazard (instantaneous risk). The language the question uses is consistent with this understanding. If I had to change it, I would say, "The death rate for smokers is twice that of *non-smokers". They also failed to mention whether these were age adjusted rates or not.

To understand this a little more deeply, relative rates and relative risks are estimated with fundamentally different models.

If you wanted to formalize a rate, you can think of this as estimating:

$$E \left( \frac{\# \mbox{events}}{\# \mbox{person-years}} \right) =\frac{\sum_i Pr(Y_i < t_i)} {\sum_i t_i} $$

($Y_i$ is the death time and $t_i$ is the observation time for the $i$-th individual, note the times are considered fixed and not random!)

You'll recognize the numerator is a bunch of CDFs, or 1-survival functions, and the relationship with survival functions and hazards is well known.

So if you took a ratio of rates:

$$ 2 = E \left( \frac{ \# \mbox{smoker deaths} \times \# \mbox{non-smoker person-years}}{\# {non-smoker deaths} \times \# \mbox{smoker person years}} \right) = \frac{\sum_i t_i}{\sum_j t_j} \frac{\sum_j Pr(Y_j < t_j)}{\sum_i Pr(Y_i < t_i)}$$

$$ = \frac{\sum_i t_i}{\sum_j t_j} \frac{ n_j-\sum_jS(t_j)}{n_i-\sum_iS(t_i)}$$

Since it's self study, you should probably do the algebra and solve the remainder of the equation!

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By "death rate", they essentially mean hazard rate. Death rates are often reported as deaths per 100,000 subjects per year, making the death rate proportional to the hazard rate, so they are not necessarily exactly alike.

In regards to the book, part "i" of the problem is quite easy as you said. Using what they ask you to prove in part "ii" of the problem, we can actually show that proportional hazards imply proportional death rates in case you don't want to just blindly take my word. Which you shouldn't.

Part "ii" says "using the relation between the hazard rates, show that the survival probability of smokers is the square of the survival probability for non-smokers" (paraphrasing).

This tells us that

$S_s(t) = S_{ns}(t)^2$

Since hazard rates are related to survival curves by

$S(t) = e^{-\int h(t)}$

This implies that

$e^{-\int h_s(t)}$ = $(e^{-\int h_{ns}(t)})^2$

Further implying that

$e^{-\int h_s(t)}$ = $(e^{-2\int h_{ns}(t)})$

Which finally leaves us with $h_s(t) = 2 h_{ns}(t)$ (provided some assumptions about the smoothness of the hazard function).

Since $h_s(t) = 2 h_{ns}(t)$, note that this has the exact same relation as described with the "death rate". Technically, this doesn't prove that "death rate" is the same as hazard rate but it implies that if death rates are proportional (whatever they are), then hazard rates are proportional.

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  • $\begingroup$ Okay, but I somehow sense that then the exercise would be a bit trivial. The level of difficulty of the other exercises is generally higher. $\endgroup$ – tomka Feb 3 '16 at 14:42
  • $\begingroup$ Please see edit above; I found some definition for death rate that is not the hazard rate. What do you think? $\endgroup$ – tomka Feb 3 '16 at 17:55
  • $\begingroup$ @ThomasKlausch: Looking at the book (thanks google books!), part "i" of the question is very easy. However, part "ii" is slightly harder. Furthermore, you can take the results that you are supposed to prove in part "ii" to prove that hazard rate = death rate. $\endgroup$ – Cliff AB Feb 3 '16 at 17:57
  • $\begingroup$ @ThomasKlausch: also, in regards to your definition, it is still the hazard rate (kind of). What's important is that when you evaluate the death rate in interval $I$, it is dependent on the size of the population in interval $I$ and thus is proportional to current living population, not total population (although in practice it's a little odder because the population gets replenished). It still works out to be basically getting the hazard rate. I had the same confusion when I first came across the terminology myself. $\endgroup$ – Cliff AB Feb 3 '16 at 18:00
  • $\begingroup$ @ThomasKlausch: ah, actually death rates are not exactly like hazard rates. I'm edited my answer (a lot) to more thoroughly answer the question. $\endgroup$ – Cliff AB Feb 3 '16 at 18:22

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