# Why Do We Need Relative Risk?

I am an MBA student that is taking courses in statistics. In our classes, we are learning about Odd's Ratios and Relative Risk.

Our Prof outlined the following examples:

• Example 1: Suppose we take a random sample of 1000 people from the general population. Of these 1000 random people, we know how many people have the disease/health and who were exposed/not exposed to some factor (e.g. cigarette smokers).

• Example 2: Suppose we take a random sample of 500 people (from the general population) who have the disease and 500 people (from the general population) who are healthy - within each group of 500 people, we know who were exposed/not exposed to some factor (e.g. cigarette smokers)

Our Prof told us that:

• In Example 1, we can use both Relative Risk and Odd's Ratio - but in Example 2, we can only use the Odd's Ratio. The logic was that Example 2 implicitly assumes that the disease rate is 50% (i.e. 50% of people in the general population have the disease) and this might not be the case.

• Although in Example 2, we can still technically calculate the Relative Risk - but this calculation would be meaningless (I am not sure why).

• In real world medical studies, if we were to pick a random sample of 1000 people from the general population, it is possible that we might only find a few people who have the disease and this will not be enough data to study the effectiveness of some drug (e.g. the drug might work, but it might be luck). This is why its often better to select predetermined samples of 500 people with the disease and 500 people without the disease to "iron out" any lucky occurrences. Since this kind of set-up does not allow us to use Relative Risk, we can now see the importance of Odd's Ratio.

If all this is true, why do we even use/learn about Relative Risk? It seems like the Odd's Ratio is more versatile (can be used in both Example 1 and Example 2) and provides similar insights as the Relative Risk. Therefore - why do we use/learn about Relative Risk?

• Only the odds ratio works in both a forward (cohort study prospectively sampled form a population) and a backwards (retrospective case-control sample where events are oversampled) fashion. And even with an all-forwards setting the OR is the measure that is capable of being constant over a wide variety of low- and high-risk subjects. Sep 11, 2022 at 18:17

Although in Example 2, we can still technically calculate the Relative Risk - but this calculation would be meaningless (I am not sure why).

Well...this reason is a bit technical but worth going through. Suppose we retrospectively sample subjects. That is to say, we first determine their outcome (cases or controls) and then determine if they were exposed (smoker or non-smoker). Let $$\phi_1$$ be the retrospective disease conditional probability of exposure given the the outcome. So $$\phi_1 = P(S \vert D)$$, where $$S$$ is smoking status for cases, and $$\phi_2 = P(S \vert \overline{D})$$ is the same for controls.

The retrospective odds ratio is

$$O R_{\text {retro }}=\frac{\phi_1 /\left(1-\phi_1\right)}{\phi_2 /\left(1-\phi_2\right)}=\frac{P(S \mid D) / P(\bar{S} \mid D)}{P(S \mid \bar{D}) / P(\bar{S} \mid \bar{D})}$$

Compare this to the prospective odds ratio

$$O R=\frac{\pi_1 /\left(1-\pi_1\right)}{\pi_2 /\left(1-\pi_2\right)}=\frac{P(D \mid S) / P(\bar{D} \mid S)}{P(D \mid \bar{S}) / P(\bar{D} \mid \bar{S})}$$

Here, $$\pi_i$$ is is the prospective risk of the outcome given smoking status. We usually want the prospective risk, it just makes more sense temporally and aligns with how we make decisions (we may decide to smoke or not given the risk of death, but never decide to die given the opportunity to smoke. See, that seems weird to even write let alone ask about).

Now, to get $$OR$$ from $$O R_{\text {retro }}$$, we need to account for the prevalence of the disease. This is because

$$P(S \vert D) \propto P(D \vert S) P(D)$$

by Bayes' rule. This means the real risk of the outcome given smoking is

$$\pi_1=P(D \mid S)=\frac{P(S\vert D) P(D)}{P(S)}=\frac{P(S \mid D) P(D)}{P(S \mid D) P(D)+P(S \mid \bar{D}) P(\bar{D})}$$

So, now we have an expression of the prospective risk in terms of the retrospective risk. The expressions are

$$\pi_1 =\frac{\phi_1 \delta}{\phi_1 \delta+\phi_2(1-\delta)}$$

$$\pi_2=\frac{\left(1-\phi_1\right) \delta}{\left(1-\phi_1\right) \delta+\left(1-\phi_2\right)(1-\delta)}$$

Where $$\delta = P(D)$$ is the prevalence of the outcome. Now, if you compute $$OR$$ using these expressions, you will find $$OR = OR_{\text{Retro}}$$

I will leave that as an exercise to the reader, it should not be too hard to do. But, if you compute the prospective relative risk $$RR$$ and the retrospective relative risk $$RR_{\text{Retro}}$$, you will find they are not equal! In order to compute the $$RR$$, you need to know $$\delta$$. If you use an odds ratio, you don't need to know $$\delta$$.

Now, if I recall from introductory epidemiology, when the disease is rare then the odds ratio is a good approximation to the risk ratio. You can probably demonstrate this by computing an appropriate limit as $$\delta \to 0^+$$. But the main point here is that

When we do a case control study we explicitly fix the proportion of cases in the sample, $$\delta$$. Doing this biases computations of the relative risk but not of the odds ratio.

Now, on to your titular question

Why do we need relative risk? [...] If all this is true, why do we even use/learn about Relative Risk?

In a word: Interpretability. Obviously, I can't say for certain, but a relative change in the risk is easier to grasp than a change in the odds. I mean, odds are hard to reason about. If I tell you your risk of death is 20% and smoking doubles that, you have a very clear sense of what that does to your risk of death.

But if I tell you your risk of death is 20% and smoking increases the odds by a factor of 2, you are probably going to interpret that as a relative risk change anyway and you'd be off by a lot in some cases(P.S. if the odds did change by a factor of 2, then your risk of death is something like 33%. You'd be off by 7 percentage points).

• @ Demetri : Is it possible to demonstrate this limit? May 31, 2023 at 18:11
• @stats_noob What limit in particular? May 31, 2023 at 18:27
• " You can probably demonstrate this by computing an appropriate limit as δ→0+" May 31, 2023 at 18:28
• would you like me to place a bounty on this question? May 31, 2023 at 18:28
• @stats_noob Why don't you write out the expression for the log odds using the formulae I've written here and compute the limit yourself? Should be tedious, but doable. May 31, 2023 at 19:24