Jtdylktuy

Let $xi_1, xi_2, xi_3$ independent random variables in $(Omega, mathcal{F},mathbb{P}).$ Also, they are evenly distributed in $[0,1]$. I need to find a distribution function of sum $xi_1+ xi_2+ xi_3$.

So, I know that $F_{xi}(x)=mathbb{P}(omega : xi(omega leq x)=mathbb{P}(xi leq x)=mathbb{P}(xi^{-1}(-infty;x])$. Also I know that $mathbb{P}(xi_1,xi_2,xi_3)=mathbb{P}(xi_1)* mathbb{P}(xi_1)* mathbb{P}(xi_1)$.

But how to use it and find a distribution function?

Convolution of two distributions.

$t_{x_0} = 0$

$t_{x_1} = 1$

$t_{h_0} = 0$

$t_{h_1} = 1$

Thus $$f_Y(t) = 0, t le t_{x_0}+t_{h_0} ,$$

$$f_Y(t) = int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(tau)f_H(t-tau)dtau, text{ } t_{x_0}+t_{h_0} le t le t_{x_1}+t_{h_1},$$

$$f_Y(t) = 0, t ge t_{x_1}+t_{h_1} ,$$

THese translate to the following solution

First convolve two uniform distributions

$X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$

$$Y(t) = x(t).h(t) = int_{-infty}^{infty} x(tau)h(t-tau)dtau$$
The above convolution reduces to

$y(t) = 0, tlt 0$

$y(t) = int_{max(0,t-1)}^{min(1,t)} dtau , 0lt t lt 2$,

$y(t) = 0 , t gt 2$

The middle one will have to split into two intervals, namely $0lt t lt 1$ and $1lt t lt 2$

$y(t) = 0, tlt 0$

$y(t) = int_{0}^{t} dtau =t, 0lt tlt 1$,

$y(t) = int_{t-1}^{1} dtau = 2-t, 1lt tlt 2$,

$y(t) = 0 , t gt 2$

Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$

For $0lt tlt 1$, the bounds are

$t_{s_0} = 0$

$t_{s_1} = 1$

$t_{y_0} = 0$

$t_{y_1} = 1$

$$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau = int_{0}^{t}tau dtau = frac{t^2}{2}, 0lt tlt 1$$,

For $1lt tlt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are

$t_{s_0} = 0$

$t_{s_1} = 1$

$t_{y_0} = 0$

$t_{y_1} = 1$

and for the interval $(1,2)$ the bounds are

$t_{s_0} = 0$

$t_{s_1} = 1$

$t_{y_0} = 1$

$t_{y_1} = 2$

Thus $$W(t) = int_{max(0,t-1)}^{min(1,t)} tau dtau + int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau$$ $$ = int_{t-1}^{1}tau dtau + int_{1}^{t}(2-tau) dtau$$ $$ = -frac{1}{2}(2t^2-6t+3), 1lt tlt 2$$,

For $2lt tlt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are

$t_{s_0} = 0$

$t_{s_1} = 1$

$t_{y_0} = 1$

$t_{y_1} = 2$

Thus $$W(t) = int_{max(1,t-1)}^{min(2,t)} (2-tau) dtau $$ $$ = int_{t-1}^{2} (2- tau) dtau$$ $$ = frac{(t-3)^2}{2}, 2lt t lt 3$$

and finally $W(t) = 0, tgt 3$

Thus the $W(t)$ is defined by

$W(t) = 0 , tlt 0$

$W(t) = frac{t^2}{2}, 0lt t lt 1$

$W(t) = -t^2+3t-frac{3}{2}, 1lt t lt 2$

$W(t) = frac{(t-3)^2}{2}, 2lt t lt 3$

$W(t) = 0 , tgt 3$

One way is (I used $Z$ instead of $xi$)$$P[Z_1+Z_2+Z_3 leq z] = int_{z_3=0}^{min(z,,1)}int_{z_2=0}^{min(z-z_3,,1)}int_{z_1=0}^{min(z-z_2-z_3,,1)}dz_1,dz_2,dz_3$$