# Uniform vs Beta(1,1) prior

Is there any difference in applying a uniform prior or a Beta(1,1) prior for your Bayesian analysis ?In which conditions is one preferred over the other ?

• No, there isn't, because the Beta(1,1) is the uniform distribution. Dec 13 '17 at 2:00
• If your parameter is constrained to lie in the interval $[0,1]$ then these two are equivalent. If your parameter can take on other values, then the Beta(1,1) prior is not a reasonable prior in the first place. Dec 13 '17 at 12:16

They both are equivalent.

$$P(\theta) = { \Gamma(\alpha + \beta) \over \Gamma(\alpha)\Gamma(\beta)} \theta^{\alpha-1}(1-\theta)^{\beta-1}$$

if $$\alpha = \beta = 1$$

$$P(\theta) = { \Gamma(\alpha + \beta) \over \Gamma(\alpha)\Gamma(\beta)} \theta^{0}(1-\theta)^{0} = {\Gamma(2) \over \Gamma(1)\Gamma(1) } = {1 \over 1} = 1$$

As you can see $$\theta| \beta=1, \alpha = 1 \sim U(0,1)$$

Because a density function identifies uniquely a distribution, and the density of a uniform in the interval $$(c=0, \ d=1)$$ is:

$$f(x) = {1\over c -d} = {1 \over 1} = 1 \quad x \in (0,1)$$

• The proof is pretty brief. It is worth mentioning that $\Gamma(1) = 1$ and $\Gamma(z + 1) = z\Gamma(z)$ for completeness, since for those not familiar with gamma functions the ratio of gammas may not look like anything similar to uniform distribution...
– Tim
Dec 13 '17 at 7:26
• @Tim Knowledge of Gamma functions is unnecessary. It suffices to point out (1) the two distributions have the same support (a crucial point that is not in evidence in this answer) and (2) when $\alpha=\beta=1,$ the Beta density is proportional to a constant on its support. That makes the two distributions identical, because both must have the same normalization constant.
– whuber
Oct 23 '18 at 14:13
• @whuber I meant more details, rather then insisting on Gamma function.
– Tim
Oct 23 '18 at 14:20
• @Tim what does "cte" stand for here? Nov 4 '19 at 12:30