I know what a conjugate prior is. But I'm confused by the name itself. Why is it called "conjugate"? A complex conjugate $z^\ast$ has a reciprocal relationship with $z$, i.e., ${z^\ast}^\ast = z$. But there isn't such a reciprocal relationship between any two elements of the triad (prior, likelihood, posterior) or at least I'm not aware of it. So why "conjugate"? Is the term overloaded?


2 Answers 2


The Oxford English Dictionary defines "conjugate" as an adjective meaning "joined together, esp. in a pair, coupled; connected, related." It's not a huge stretch to imagine that a conjugate prior has a special and strong connection to its posterior.

It's used in a similar sense in chemistry (conjugate acid/base; conjugate solution), botany (leaves that grow in pairs, especially when there's only one pair), optics (conjugate foci), and linguistics (conjugations are forms of the same root word).

While some have a "reciprocal" implication, others don't, so I don't think it's a necessary element of the meaning.

Wikipedia credits Raiffa and Schlaifer for coining the term (annoyingly, it's not in the OED). Here's the first mention of it in their 1961 book, which seems to be using the "joined" sense of conjugate.

Screenshot of preface from Raiffa and Schlaifer

We show that whenever (1) any possible experimental outcome can be described by a sufficient statistic of fixed dimensionality (i.e., an $s$-tuple $(y_1, y_2, \ldots y_s)$ where $s$ does not depend on the "size" of the experiment), and 2) the likelihood of every outcome is given by a reasonably simple formula with $y_1, y_2, \ldots y_s$ as its arguments, we can obtain a very tractable family of "conjugate" prior distributions simply by interchanging the roles of variables and parameters in the algebraic expression for the sample likelihood, and the posterior distribution will be a member of the same family as the prior. "

  • 1
    $\begingroup$ Very good point, this is also the earliest reference I am aware of. I would presume they used "conjugate" in a vague meaning of "associated" rather than more mathematically. $\endgroup$
    – Xi'an
    Jul 25, 2020 at 14:46
  • $\begingroup$ I think that's actually the original meaning, with the "reciprocal"/ mathematical sense coming later as a special case. After all, one obvious way for things to be paired is for one to be the reciprocal of the other. $\endgroup$ Jul 25, 2020 at 16:51

I believe the origin is somehow related to the following concepts:

  • eigenvector: a vector $ \mathbf{x} $ is called an eigenvector of a matrix $ \mathbf{A} $ if $ \mathbf{A}\mathbf{x} $ = $k\mathbf{x}$ , meaning $ \mathbf{A}\mathbf{x} $ has the same form as $ \mathbf{x} $ (just different by a scaling factor $k$ called eigenvalue of $ \mathbf{A} $), hope you start to see this is the same logic with conjugate prior.

  • eigenfunction: see this analogy between conjugate prior and eigenfunction. The concept of eigenvector is extended to functions in Functional Analysis. Given a linear transformation $L$ (eg. a differential or integral operator), its eigenfunctions are functions $f$ such that $Lf$ is simply $kf$, ie. $f$ scaled by a scalar. The eigenfunctions are very useful in solving differential equations as they provide us with very convenient representation of their solutions. These are also related to Fourier transforms, where eigenfunctions of a Fourier transform are sine and cosine functions. In fact, it can be proved that any periodical function can be approximated as a linear combination of sine and cosine functions. Also, Fourier transform of a Gaussian function is another Gaussian function, again same logic with conjugated prior.

  • 3
    $\begingroup$ can you please elaborate what you mean by the usage of conjugated of a function in this context. $\endgroup$ Jul 25, 2020 at 12:42
  • 4
    $\begingroup$ I frankly fail to see the connection between eigenfunctions and conjugate priors. $\endgroup$
    – Xi'an
    Jul 25, 2020 at 14:50
  • $\begingroup$ Hi @Xi'an, just edited to clarify the connection between the two concepts. $\endgroup$
    – Victor Luu
    Jul 25, 2020 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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