2
$\begingroup$

The NYT reports:

The former F.B.I. director and his deputy, both of whom former President Donald J. Trump wanted prosecuted, were selected for a rare audit program that the tax agency says is random.

The odds of being selected for that audit in any given year are tiny — out of nearly 153 million individual returns filed for 2017, for example, the I.R.S. targeted about 5,000, or roughly one out of 30,600.

I'm trying to understand how we can know if these are independent events -- just in terms of the statistics, without knowing anything more about the case.

If we say independence means: P(XY) - (P(X) + P(Y)) = 0

Then we are saying: the probability of the joint occurrence is equivalent to the linear addition of the individual probabilities.

So is this true?

Let's say the process of selection is not random but is instead corrupt and the two have not yet been selected for tax audit. It's 2017, and the probability of joint occurrence (P(XY)) is high, say 95%.

On the face of it, the probabilities do not seem independent. However, maybe they are. We could say that the probability that one gets audited is 1/30,600 (P(X)) and the other is also 1/30,600 (P(Y)).

That implies dependence by our definition, even though this doesn't intuitively make sense to me. Is it correct to think about the joint probabilities this way?

It seems intuitively that once you know that one is selected, the odds the other is selected goes up massively. So there's some dependence.

But should we simply consider the individual probabilities so low -- or should they also be much higher because each person has a high chance of being audited, jointly or not? I'm confused how we rate individual probabilities in this instance and how it connects to independence.

$\endgroup$
3
  • 1
    $\begingroup$ You left out the part where they were picked in different draws, which changes the question ("how likely is it that two people are ever both picked for audit?", similar to birthday paradox). Also, "random" is not necessarily uniformly random, and "corruption" is not the only alternative. This is some kind of "special" deep audit and the article does not say that the returns chosen are selected from the entire pool of returns; they could be pre-selected according to some features which, for two people working essentially the same job at the same employer, could be highly correlated. $\endgroup$
    – Chris Haug
    Jul 7, 2022 at 10:59
  • $\begingroup$ @ChrisHaug Yes, fair enough. I simplified heavily for my example :) $\endgroup$ Jul 7, 2022 at 13:22
  • 1
    $\begingroup$ I asked a related question at stats.stackexchange.com/questions/581157/… in case you're interested $\endgroup$
    – Adrian
    Jul 8, 2022 at 16:12

1 Answer 1

1
$\begingroup$

If two events $X$ and $Y$ are independent then $$ P(X \cap Y) = P(X)P(Y), $$ where $X\cap Y$ is the event that both $X$ and $Y$ do happen.

IIUC, you want to know how likely it is to have both being selected for the audit provided the selection is completely random (i.e. in particular independent) and $P(X) = P(Y) = 1/30,600$. According to the above formula, that probability is: $$ P(X\cap Y) = P(X)P(Y) = \frac{1}{30,600}\cdot \frac{1}{30,600} = \frac{1}{936,360,000} \approx 0.000000001. $$

By the way, if the selection of both is independent, then knowing that one is selected will not change the probability that the other one is selected, that is the very meaning of independence.

$\endgroup$
3
  • $\begingroup$ I guess it doesn't seem like they are random. So I'm wondering how to estimate the individual and joint probabilities in that case. $\endgroup$ Jul 7, 2022 at 10:07
  • 1
    $\begingroup$ @Learningstatsbyexample The individual probability doesn't change, it is still $1/30,600$. But without independence, the product formula doesn't hold anymore. Rather, $P(X\cap Y) = P(X|Y)P(Y)$, and $P(X|Y)$ is probably rather difficult to come by. $\endgroup$
    – frank
    Jul 7, 2022 at 10:56
  • 1
    $\begingroup$ This is a great follow-up.. nytimes.com/2022/07/07/upshot/comey-mccabe-tax-audits.html $\endgroup$ Jul 7, 2022 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.