From another Cross Validated question

Given $X\sim N(\mu_X, \sigma_X^2)$ and $Y\sim N(\mu_Y, \sigma_Y^2)$ are independent, and you know $X+Y=s$. What is the expected value of $X$?

The question makes it sound like $Y=s-X$ yet $X$ and $Y$ are independent. How can this be?

$$cov(X,Y)\\=cov(X,s-X)\\=-var(X)\ne 0$$

Nonzero covariance is one form of dependence and precludes independence.

  • $\begingroup$ See if the image at stats.stackexchange.com/a/9073/2958 helps $\endgroup$
    – Henry
    Aug 18, 2022 at 0:46
  • $\begingroup$ It isn't that the RVs are deterministically adding to $s$, rather the question is asking about conditioning on the event. $\endgroup$
    – Ben
    Aug 18, 2022 at 0:46
  • $\begingroup$ @Ben That seems to be the approach taken in the other question, not an assumption of the problem faced in the interview. $\endgroup$
    – Dave
    Aug 18, 2022 at 0:49
  • $\begingroup$ I agree with @Ben and have interpreted the questions as asking for the conditional expectation of $X$ given $X+Y=s$ for a specific (realized) $s$. $\endgroup$
    – statmerkur
    Aug 18, 2022 at 0:56
  • $\begingroup$ Hi @Dave. Writing mathematical expressions in plain English necessarily leads to imprecision. "If $X+Y=s$, what is the expectation of $X$" is just slightly sloppy wording for a conditional expectation. $\endgroup$
    – Ben
    Aug 18, 2022 at 1:56

2 Answers 2


Ben's comment pretty much says it all.

$X, Y$ being independent doesn't mean there cannot be an event where their added realisations have a particular aggregate, say $s.$ But the distinction must be noted: $$Y :\ne X-s;$$ the necessary condition of independence is uncorrelatedness, i.e. $\mathbb{Cov}(X, Y) = 0,$ but in no way $Y$ is defined to be $X- s. $


This sounds a bit like the idea behind collider bias, the reverse of confounding bias.

  • Confounding bias: x and y are found to be correlated because both are caused by z.
  • Collider bias: x and y are found to be correlated because they both cause z and there is a selection effect based on z.

example of bias

Image from the question:Can spurious correlations exist in the (theoretical) population?

So $X$ and $Y$ may not be correlated, but when you condition on $s$ then they are correlated.

It is written a bit implicitly in the interview question but the statement implies a conditioning

Given that ... and you know $X+Y=s$

You might assume that this means a particular sampled individual case has $X+Y=s$. Otherwise there is a contradiction. $X$ and $Y$ can not be independent while also $X+Y=s$.


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