Prove the double angle formula

Question

Let $X$ and $Y$ be independent random variables, where $X\in U(-1,1)$ and $Y$ assumes the values of $+1$ and $-1$ with probabilities $1/2$.

Show first that $Z=X+Y = U(-2,2)$ by finding the distribution function of Z.

Then translate that fact into a statement about the corresponding characteristic functions.

Finally show that $\sin(2t) = 2 \sin(t) \cos(t)$

Answer 1

PDF of $U(-1,1)$ is $$ f_X(x) = \left \{ \begin{array}{cl} \frac 12 & -1 \le x \le 1 \\ 0 & \text{else} \end{array}\right . $$ PMF (Probability mass function) of second RV $$ f_Y'(y) = \left \{ \begin{array}{cl} \frac 12 & y \in \{-1,1\} \\ 0 & \text{else} \end{array}\right . $$ I put $'$ to the $f_Y$ to distinguish it from actual PDF of $Y$ which is given below.

Using generalized functions concept (Distribution) you can rewrite them as $$ f_X(x) = \frac 12\left [\theta(x+1)-\theta(x-1)\right] \\ f_Y(y) = \frac 12 \left [ \delta(y+1) + \delta(y-1)\right] $$ where $\theta(x)$ is Heaviside step function and $\delta(x)$ Dirac delta function.

If you have $Z = X+Y$ where $X$ and $Y$ are independent, then $$ f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y)dy $$ which is a convolution of $f_X$ and $f_Y$ w.r.t variable $y$. $$ f_Z(z) = \frac 14 \int_{-\infty}^\infty \left [ \theta(z-y+1)-\theta(z-y-1)\right ] \cdot \left [\delta(y+1) + \delta(y-1 \right ]dy = \\ = \frac 14 \left \{ \int_{-\infty}^\infty \left [\theta (z-y+1) - \theta(z-y-1)\right ] \delta(y+1)dy \right \} + \\ \frac 14 \left \{ \int_{-\infty}^\infty \left [\theta(z-y+1)- \theta(z-y-1)\right ]\delta(y-1)dy \right \} = $$ If you use property of the Dirac delta function $$ \int_{-\infty}^\infty f(t-\tau) \delta(\tau)d\tau = f(t) $$ you can state integral as $$ \frac 14 \left [ \theta(z+2) - \theta(z)\right ] + \frac 14 \left [ \theta(z)-\theta(z-2) \right ] = \frac 14 \left[ \theta(z+2)-\theta(z-2)\right ] $$ which is PDF of the $U(-2,2)$.

So $$ f_Z(z) = \left \{\begin{array}{cl} \frac 14 & -2 \le z \le 2 \\ 0 & \text{else} \end{array}\right . $$

PS

I'm not sure I understood second part about double angle, since that identity is always true, regardless whether argument is some number or RV.

Answer 2

There is a $\frac12$ chance that $Y=1$ in which case the distribution of $X+Y$ is $$ f_1(x)=\left\{\begin{align} \tfrac12\qquad&\text{if }x\in[0,2]\\ 0\qquad&\text{otherwise} \end{align}\right. $$ There is a $\frac12$ chance that $Y=-1$ in which case the distribution of $X+Y$ is $$ f_{-1}(x)=\left\{\begin{align} \tfrac12\qquad&\text{if }x\in[-2,0]\\ 0\qquad&\text{otherwise} \end{align}\right. $$ Thus, the distribution of $X+Y$ is $$ \frac12f_{-1}(x)+\frac12f_1(x)=\left\{\begin{align} \tfrac14\qquad&\text{if }x\in[-2,0]\\ \tfrac14\qquad&\text{if }x\in[0,2]\\ 0\qquad&\text{otherwise} \end{align}\right. $$ which is the convolution of the distributions of $X$ and $Y$. That is, the distribution is $U(-2,2)$.

I don't see a relation between this and the double angle formula for $\sin$.

Answer 3

For the double angle formula:

Using $sin(2t) = sin(t+t)$, apply the sum formula for sine: $sin(a+b) = sin(a)cos(b) + cos(a)sin(b)$