Suppose I flip a coin 10 times and get 7 heads - based on a Binomial Distribution, I can create a Likelihood Function and estimate the probability of getting a head is 0.7.

Now, suppose someone tells me that they have very strong evidence to believe that the probability of getting a head is 0.9. Suppose I decide to use a Beta Distribution(alpha, beta) as a prior.

My Question: I know that the Expected Value of the Beta Distribution is 'alpha/(alpha + beta)'. If I want to use the Prior Information that was provided to me - does this mean that I have choose values of 'alpha and beta' such that 'alpha/(alpha+beta)' = 0.9? For example - based on the n = 10 samples I observed, in this case, would alpha = 9 and beta = 1?


3 Answers 3


does this mean that I have choose values of 'alpha and beta' such that 'alpha/(alpha+beta)' = 0.9?

That's correct, but the actual values you choose determine how strong your prior is.

In this particular setup, there's an intuitive explanation of the Beta prior: it is equivalent to having previously observed $\alpha$ heads and $\beta$ tails from the same coin.

You can thus see that $(\alpha, \beta)=(9,1)$ is much weaker than $(\alpha, \beta)=(900000,100000)$.


If I understand correctly, your problem is not defined correctly:

  • The likelihood is a Binomial distribution (you flip the coin)
  • The Prior is a Beta distribution (representing the distribution of the parameter p of the Binomial)
  • Beta is a conjugate prior for Binomial, it means the posterior is still a Beta distribution.

Here you can find the following details: enter image description here

Therefore, in order to identify the "believe", that is a Beta distributions, you need 2 parameters:

  • alpha & beta (your problem is solved)


  • mean & variance, in order to solve this equation:

$$\mathbf E[X]=\frac\alpha{\alpha+\beta}$$

$$\mathbf V[X]= \frac{\alpha\beta}{(\alpha+\beta) ^2(\alpha+\beta+1) }$$

as reported here: Calculating the parameters of a Beta distribution using the mean and variance.

Therefore: Variance is missing.

  • $\begingroup$ ... and OP only specified a single moment, the mean. Welcome to CrossValidated! $\endgroup$ Feb 1 at 9:44
  • 1
    $\begingroup$ Welcome to Cross Validated. This is a MathJax enabled site: Instructions on how to use LaTeX on CrossValidated. Please use LaTex to type equations as it makes for more readable answers. $\endgroup$
    – dipetkov
    Feb 1 at 10:59

The typical approach in this case would be to use the conjugate prior (the beta distribution) for the unknown probability parameter and parameterise this using a stipulated prior mean $0 \leqslant \mu_0 \leqslant 1$ and prior effective-sample-size $n_0 \geqslant 0$. You can then use the reparameterisation of the beta distribution with parameters:

$$\begin{align} \alpha &= \mu_0 (n_0+2), \\[6pt] \beta &= (1-\mu_0) (n_0+2). \\[6pt] \end{align}$$

From the information you have given you would choose $\mu_0 = 0.9$ and you would then have to choose an appropriate value for the prior effective sample-size. This is a measure of the strength of your prior belief --- the higher the effective-sample-size the stronger your prior belief (and so the lower the prior variance). You describe the prior belief as "very strong" so you could choose a value like $n_0 = 100$ to represent this, which would mean that you consider your prior belief to be as strong as if it were formed from one-hundred observed data points.


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