<p>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 $\frac{1}{2\pi}$.</p>

<p>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 $4\pi$.</p>

<p>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 $\frac{1}{\pi}$.</p>

CC BY-SA 3.0
2017-05-05T10:06:20.270Z
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
CC BY-SA 3.0
2017-05-08T08:56:28.300Z
I think Olivier's answer explains that well. Steradians and the unit sphere go together in the same way that radians and the unit circle go together. If you change units so that your sphere has a different radius, that radius becomes an extra factor in the integration. "Choosing a direction" and "choosing a point on the surface of a unit sphere" can be treated as the same thing.
Dan Hulme
CC BY-SA 3.0
2017-05-02T15:52:36.140Z
2017-05-02T15:52:36.140Z
Dan Hulme
<p>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.</p>

CC BY-SA 3.0
2017-05-02T16:38:40.207Z
2017-05-02T17:49:07.707Z
2017-05-02T17:49:07.707Z
Olivier
<p>I'm trying to figure out why the probability density function for picking a direction uniformly on the hemisphere is $\frac{1}{2\pi}$.</p>

<p>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.</p>

<p>EDIT: I understand this now. My confusion arose because I wasn't thinking clearly about the <strong>units</strong> 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)!</p>

CC BY-SA 3.0
2017-05-02T14:55:36.033Z
IIRC if you integrate the PDF over the entire domain the result must be 1.
ratchet freak
<mathematics>
Why is $pdf(\Psi) = \frac{1}{2\pi}$ when picking from hemisphere uniformly?
PeteUK
# Why is $pdf(\Psi) = \frac{1}{2\pi}$ when picking from hemisphere uniformly?

(Asked by: username_0 on 2017-05-02)

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)!

**Tags:** mathematics

> **Comments:**
> **username_1**: IIRC if you integrate the PDF over the entire domain the result must be 1.  

---

## 2 answers

### Answer by username_3 on 2017-05-02. Score: 4 | Accepted answer

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.


---

### Answer by username_2 on 2017-05-02. Score: 4

It's like this because the surface area of a unit sphere is $4\pi$. As username_1 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 $\frac{1}{\pi}$.

> **Comments:**
> **username_0**: 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?  
> **username_2**: I think username_3's answer explains that well. Steradians and the unit sphere go together in the same way that radians and the unit circle go together. If you change units so that your sphere has a different radius, that radius becomes an extra factor in the integration. "Choosing a direction" and "choosing a point on the surface of a unit sphere" can be treated as the same thing.  

---

