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I would be really grateful if someone could check what I have done here; it should be quick:

Let $\Phi$ be a random variable taking values in $[0,\pi]$ with PDF $f(\phi)=\frac{1}{2}\sin\phi$. Define: $$h(\phi)=a\cos(\phi)+b$$ where $a,b$ are positive constants. I need to find the distribution of $h(\Phi)$.

Attempt:

The PDF of $h(\Phi)$ should be: $$f(h^{-1}(\phi))\cdot \frac{d}{d\phi}h^{-1}(\phi)$$ I get this to be the constant $-1/(2a)$, so $h(\Phi)$ has uniform distribution. In particular

$$h(\Phi)\sim \mathcal{U}\left(b-a,b+a\right)$$

Can I check that this is correct?

user118224
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1 Answers1

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The computation is correct. I will add that this is an instance of a more general fact: if $\Phi$ is a random variable with continuous CDF $g$, then $g\circ \Phi$ is uniformly distributed in $[0,1]$ (see here).

In your case, the CDF of $\phi$ is $\frac12(1-\cos\phi)$, so applying this function to $\Phi$ produces a variable $\sim U(0,1)$. And $a\cos\phi+b$ is just a linear transformation of $\frac12(1-\cos\phi)$, scaling and adding a constant. More precisely, it is the transformation $x\mapsto -2a x+(a+b)$, hence the interval $[0,1]$ is mapped to $[b-a, b+a]$.

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