# Why is $pdf(\Psi) = \frac{1}{2\pi}$ when picking from hemisphere uniformly?

I'm trying to figure out why the probability density function for picking a direction uniformly on the hemisphere is $\frac{1}{2\pi}$.

Something tells me this is related to the number of steradians in a hemisphere being $2\pi$, but I can't see how this is connected. We could come up with another measure for solid angle ("blibs" for example) where the number of blibs in the hemisphere would be $8000\pi$, but I'd expect the probability for picking a direction uniformly to be unchanged.

EDIT: I understand this now. My confusion arose because I wasn't thinking clearly about the units of the probability density function, i.e. $\frac{1}{sr}$. I got thrown off by thinking of area, and confusingly thought the pdf units were inverse area (or something)!

• IIRC if you integrate the PDF over the entire domain the result must be 1. – ratchet freak May 2 '17 at 14:55

Actually, it would change if you changed units in the $\textit{pdf}$ definition. The fundamental reason is that the $\textit{pdf}$ is defined as the probability per steradian. That's what the density part means. You could very well redefine it as the probability per hemisphere and end up with a $\textit{pdf}$ of $1$ for your example.
It's like this because the surface area of a unit sphere is $4\pi$. As ratchet freak points out, the integral of a probability distribution over its domain has to be 1. Put another way, the probability of choosing some direction is 1. The surface area of the hemisphere is $2\pi$. The constant that you integrate over a $2\pi$ domain to get 1 is $\frac{1}{2\pi}$.
The connection to steradians is that 1 steradian is defined as the solid angle that subtends a surface of area 1 on a unit sphere. That is, a unit sphere has $4\pi$ steradians and a surface area of $4\pi$.
It's just the same in two dimensions. A radian is chosen so that a 1 radian arc of a unit circle has length 1, so there are $2\pi$ radians in a circle, and $\pi$ radians in a semicircle. If you choose 2D directions uniformly on a semicircle, the probability distribution is $\frac{1}{\pi}$.
• I wasn't think in terms of area, so thanks a lot. I still have an unresolved paradox in my poor brain. Imagine a unit hemisphere and a hemisphere of radius 2, centred at the same point. The first has surface area $2\pi$ and the second $8\pi$. The probability of picking a particular direction over these hemispheres seems the same, but it's going to be $\frac{1}{2\pi}$ and $\frac{1}{8\pi}$ respectively. I can't get my head around why it's different mathematically but seemingly the same theoretically. Put another way, why does the unit hemisphere get precedence over ones of other radii? – PeteUK May 5 '17 at 10:06