Does this NYT article incorrectly assume independent increments?

Question

The article plots for every 100 women that use a certain type of contraception method the number of unplanned pregnancies over time.

https://www.nytimes.com/interactive/2014/09/14/sunday-review/unplanned-pregnancies.html?_r=0

In particular at the end of the article they say:

The numbers are calculated as follows:

$ \mathbb P(\text{Not pregnant after year N}) = \mathbb P(\text{Not pregnant after year 1})^N$

Indeed, success rate of contraception is probability not pregnant in year 1. See e.g. https://www.cdc.gov/reproductivehealth/contraception/unintendedpregnancy/pdf/contraceptive_methods_508.pdf

This is true if probability pregnant in a year is independent of the year before, but seems highly unlikely to be true. If you use contraception the wrong way, it will probably go wrong in the first year, and if it does not, then it probably won't go wrong the year after?

Answer 1

Sorry, I cannot agree with the independence assumption. Fertility in women, even without contraception is a function of age, such that, without contraception the

Chances of getting pregnant without IVF (in vitro fertilization)

Starting at about age 32, a woman’s chances of conceiving decrease gradually but significantly.
From age 35, the fertility decline speeds up.
By age 40, fertility has fallen by half.
At 30, the chance of conceiving each month is about 20%. At 40 it’s around 5%.
Note (mine) after age ~49 menopause occurs and when it does, women are infertile.

The rate of pregnancy is also a function of frequency of intercourse, which also changes with age:

About 5% of single women between the ages of 18 and 24 had sex 4 or more times per week, but 24% of married women did.
Like the men, just under half of the women between the ages of 25 and 59 had sex a few times per month to weekly, more than their single and partnered peers.
Sexual frequency did decrease with age for women, although almost a quarter of partnered women over age 70 had sex more than 4 times a week.

Relative time of ovulation, intercourse, and female age:

Finally, to consider the effectiveness of contraception on an annualized basis, one must consider not only decreasing fertility and variable but generally somewhat decreasing sexual frequency with age, but likely also myriad other factors. For example, the percentage of women who are postpartum increases with age, and postpartal women may have a different contraceptive usage effectiveness than the nulliparous, age of the partner at time of intercourse relative to ovulation, see image:

timing of intercourse relative to ovulation, having a huge impact on fertility, also reflects on the likelihood of pregnancy even when other factors, like contraception is considered. Thus, a woman who relies on the rhythm method, as well as one or more other methods of contraception, i.e., a woman who both knows her body functions, and uses that knowledge (and as knowledge is acquired) may eventually do an increasingly effective job of avoiding pregnancy, such that there is essentially no chance for independence of fertility with elapsed age.

Answer 2

Here is an account of the probabilities relevant to the problem at hand.

Denote by $z_n$ the event 'no pregnancy after $n$ years' for a woman using some type of contraception. Then $$ P(z_N) = P(z_N | z_{N-1}) P(z_{N-1} | z_{N-2})\cdots P(z_2 | z_1)P(z_1). $$ The problem is that the NYT asumes $P(z_i | z_{i-1}) = P(z_1) = p$, for all $i$, while knowing $z_{i-1}$ may provide evidence that the woman makes good use of the contraception method and may be experienced with it. We should therefore expect that $$ P(z_N | z_{N-1}) > P(z_{N-1} | z_{N-2}) > \cdots > P(z_1). $$ This implies $$ P(\text{'at least 1 pregnancy after $N$ years'}) < 1-p^n $$ rather than the equality claimed by the NYT.

Addendum. (Alternative presentation of user385948's answer)

Every woman $w$ using a particular type of contraceptive, among $M$ other woman, has her own fixed probability $q_w$ of not getting an unwanted pregnancy in a year. The average success rate of the contraceptive over one year is $ p =\tfrac{1}{M}\sum_w q_w$. The average success rate after $N$ years, assuming year-to-year independence, is $\tfrac{1}{M}\sum_w q_w^N$. However, $$ \tfrac{1}{M}\sum_w q_w^N \geq \left(\tfrac{1}{M}\sum_w q_w\right)^N = p^N, $$ by Jensen's inequality, with equality if and only if $q_w$ is constant over $w$.

Therefore, in average, generally stricly less than $(1-p^{10})\times 100$ woman in a hundred will have an unwanted pregnancy over a period of 10 years.

Answer 3

From a probabilist perspective I would expect that $$ \mathbb P(\text{Not pregnant after year N}) \geq \mathbb P(\text{Not pregnant after year 1})^N.$$ This expectation is motivated as follows. Assume that at time $t=0$, every woman is assigned a (potentially different) number $p\in[0,1]$, the probability that she will get pregnant in the first year. If she did not get pregnant after $k$ years, then the probability that she gets pregnant in the $k+1$-th year is again $p$. Then $1-\mathbb{E}p$ is exactly $$\mathbb P(\text{Not pregnant after year 1}).$$ We want to prove that $$\mathbb P(\text{Not pregnant after year 1})^N$$ is a lower bound for $$\mathbb P(\text{Not pregnant after year N}).$$ But, given the number of women and $$1-\mathbb{E}p=\mathbb P(\text{Not pregnant after year 1}),$$ we can optimise the values for $p$ of the individual women to minimise $$\mathbb P(\text{Not pregnant after year N}).$$ There is one global minimum, it is "assign $p' =\mathbb{E} p$ to any woman" (so $p'$ is deterministic), and for this minimum we have equality (because indeed everything is independent). The inequality then follows. To illustrate this with an example, suppose that we have two women, having $p=0$ and $p=1$. Then $$\frac{1}{2}=\mathbb P(\text{Not pregnant after year N})>\mathbb P(\text{Not pregnant after year 1})^N =\frac{1}{2^N}$$ for $N>1$.