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Given random variable $x$, which has a bounded $\arcsin(a,b)$ distribution. I am trying to find the distribution $Y$ such that $y=\arccos(x)$.

My intuition tells me that $Y$ is an uniform$(\arccos(a),\arccos(b))$ distribution. How do I go about proving this? I have been trying transformation of random variables, but the function nor it's derivative are monotonic and single-valued so I'm not sure it applies.

Thanks in advance.

Anton Vrdoljak
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moddinati
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1 Answers1

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The Claim is not valid. For example let $$X\sim ArcSin(5,8)$$. Arcsine_distribution

so $5<X<8$, so It claimed $$Arccos(X)\sim Uniform(ArcCos(5),ArcCos(8))$$

What is $ArcCos(5)$???!!!

let we choose $a$ and $b$ such that $-1<b*sin(U)+a<1$ and so $ArcCos(b*sin(U)+a)$ is defined. Folllowing R code show $ArcCos(b*sin(U)+a)$ does not follow uniform

 set.seed(1)
 U<-runif(100000,-pi,pi)
 X<-.1*sin(U)+.2
 plot(density(acos(X)))

enter image description here Hint: I think if $U\sim Uniform(-\pi,\pi)$ so $ArcCos(sin(U))\sim Uniform$.

I think the key is behind on the following relation

$$Arcsin(x)+Arccos(x)=\frac{\pi}{2}$$

Relationships_between_trigonometric_functions_and_inverse_trigonometric_functions

$U\sim Uniform(-\pi,\pi)$ so $$X=\sin(U)\sim ArcSin(-1,1)$$ and $$Arcsin(X)\sim Uniform(-\pi, \pi)$$

Arcsine_distribution#Related_distributions

so $$Arccos(X)=\frac{\pi}{2} -Arcsin(X)\sim ??$$

 set.seed(1)
 U<-runif(100000,-pi,pi)
 X<-sin(U)
 plot(density(acos(X)))

enter image description here

Masoud
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  • Ok thanks. Not sure if I'm fully following. Using your variables above, let's define $Y=\beta X + \alpha$ ~ $ArcSin(\alpha-\beta,\alpha+\beta)$. Therefore, my problem would be $$ArcCos(Y) = \frac{\pi}{2} - ArcSin(Y) = \frac{\pi}{2} - Arcsin(\beta X + \alpha) = \frac{\pi}{2} - Arcsin(\beta sin(U) + \alpha)$$ Still not sure how to go from here? – moddinati Mar 16 '20 at 22:07
  • $ArcCos(Y)$ and $Arcsin(Y)$ are defined when $-1\leq Y\leq 1$.what do you mean when $ArcSin(\beta \sin(U) +\alpha)$. for example when $u=0$ and $\alpha=5$ , what does $ArcSin(\beta \sin(u) +\alpha)$ mean? – Masoud Mar 17 '20 at 11:05