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I was learning about Bayesian Parameter Estimation when I came across Uniform prior. I saw the answers to this question for some insight into it. In the source I'm referring to, this is how the function is described: enter image description here

Doesn't this suggest that for any θ, P(θ) = 1? How is that possible? Have I misunderstood the description?

Any insight is appreciated!

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    $\begingroup$ Welcome to Cross Validated! I suggest drawing out that function, say on $(-1,2)$. Does that look like a uniform distribution to you? $\endgroup$
    – Dave
    Commented Apr 12, 2022 at 17:30
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    $\begingroup$ You can see a plot of $P$ in the second figure of my post at stats.stackexchange.com/a/43075/919. The first figure plots the corresponding distribution function (its CDF). Reading between the lines, I would guess you might be misinterpreting $P$ as giving probabilities. It doesn't: it's a density. For the distinction, see stats.stackexchange.com/questions/4220. In fact, the probability of any single value of $\theta$ is zero. See stats.stackexchange.com/questions/142730. $\endgroup$
    – whuber
    Commented Apr 12, 2022 at 19:36
  • $\begingroup$ Can you expand on your background? Like, how much you know about the differences between probability distribution, density function, cdf... $\endgroup$
    – Xi'an
    Commented Apr 13, 2022 at 5:41

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This $P(\theta)$ is the probability density, therefore $$ \text{Probability of finding } a \leq \theta \leq b = \int_a^b d\theta P(\theta). $$

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  • $\begingroup$ I don't really see how this answers the question... $\endgroup$
    – jbowman
    Commented Apr 12, 2022 at 19:53
  • $\begingroup$ OP asked how is it possible for any $\theta\in[0,1]$ the $P(\theta)$ to be 1. That's why I think she/he took it as a probability instead of probability density. $\endgroup$
    – Peter Pang
    Commented Apr 12, 2022 at 20:59
  • $\begingroup$ I can see that, OK. $\endgroup$
    – jbowman
    Commented Apr 12, 2022 at 22:47

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