Yeah, but the convolution of two distributions is the sum of their characteristic functions, i.e., the Fourier transform of their PDFs. Mathematica gave me a nice solution for this today. For Z = \cos \Theta_1 + \cos \Theta_2:
f_Z(z)=\frac{K(1-\frac{z^2}{4})}{\pi^2}
Where K(k) is the...
I am trying to derive the distribution for the sum of two random vectors, such that:
\begin{align}
X &= L_1 \cos \Theta_1 + L_2 \cos \Theta_2 \\
Y &= L_1 \sin \Theta_1 + L_2 \sin \Theta_2
\end{align}
With:
\begin{align}
L_1 &\sim \mathcal{U}(0,m_1) \\
L_2 &\sim...
Great, I see it now! Finally, I have:
\begin{align}
F_Y(y) &= \frac{\pi - \cos^{-1}{y}}{\pi} \\
&= \frac{\frac{\pi}{2} + \sin^{-1}{y}}{\pi}
\end{align}
Which gives me:
f_Y(y) = \frac{1}{\pi \sqrt{1 - y^2}}
Thanks so much for all of your help and patience.
I'm not sure- I still feel like I'm missing something simple. Looking at the sin plot I can see a geometric solution as follows:
A_1=y(\pi-2\sin^{-1}y)
\begin{align}
A_2&=2\int^{\sin^{-1}y}_0 \sin x \,dx \\
&=2\left[ \cos 0 - \cos(\sin^{-1} y) \right] \\
&=2\left( 1 -...
Thanks for all of your replies. All of your points make sense, but I'm still not sure how to proceed.
It's obvious that:
\begin{align}
P(X\le1)&=1 \\
P(X\le0)&=0.5 \\
P(X\le-1)&=0 \\
\end{align}
It's also clear that the CDF will be the same for both sin(Y) and cos(Y) since...
Sorry that it is unclear. I have a random variable Y which is uniformly distributed, and I am interested in the PDF of X, given that:
X = cos(Y)
Does that clarify?
Homework Statement
I want to find the PDF for arccos and arcsin of a uniform random number. Given:
Y\sim\mathcal{U}(0,2\pi) \\
X = cos(Y)
The Attempt at a Solution
I started with trying to find the CDF:
\begin{align}
F_X& = P(X \le x) \\
& = P(cos(Y) \le x) \\
& = P(Y \le arccos(x))...