Recent content by thapyhap

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...

Y is uniformly distributed between [0, 2∏)

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))...