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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?

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Convolution of two distributions.

$t_{x_0} = 0$

$t_{x_1} = 1$

$t_{h_0} = 0$

$t_{h_1} = 1$

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

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

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

THese translate to the following solution

First convolve two uniform distributions

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

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

$y(t) = 0, t\lt 0$

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

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

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

$y(t) = 0, t\lt 0$

$y(t) = \int_{0}^{t} d\tau =t, 0\lt t\lt 1$,

$y(t) = \int_{t-1}^{1} d\tau = 2-t, 1\lt t\lt 2$,

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

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

For $0\lt t\lt 1$, the bounds are

$t_{s_0} = 0$

$t_{s_1} = 1$

$t_{y_0} = 0$

$t_{y_1} = 1$

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

For $1\lt t\lt 2$, $S(t)$ 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 $$W(t) = \int_{max(0,t-1)}^{min(1,t)} \tau d\tau + \int_{max(1,t-1)}^{min(2,t)} (2-\tau) d\tau$$ $$ = \int_{t-1}^{1}\tau d\tau + \int_{1}^{t}(2-\tau) d\tau$$ $$ = -\frac{1}{2}(2t^2-6t+3), 1\lt t\lt 2$$,

For $2\lt t\lt 3$, $S(t)$ 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 $$W(t) = \int_{max(1,t-1)}^{min(2,t)} (2-\tau) d\tau $$ $$ = \int_{t-1}^{2} (2- \tau) d\tau$$ $$ = \frac{(t-3)^2}{2}, 2\lt t \lt 3$$

and finally $W(t) = 0, t\gt 3$

Thus the $W(t)$ is defined by

$W(t) = 0 , t\lt 0$

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

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

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

$W(t) = 0 , t\gt 3$

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

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  • $\begingroup$ How to integrate this? $\endgroup$ – Atstovas Dec 10 '18 at 9:00
  • $\begingroup$ I guess you need to find the possible cases. The simplest one is when $z < 1$. In this case, the upper limits are $z, \,z-z_3,\,z-z_2-z_3$. For $z\geq 1$ there are more cases. $\endgroup$ – BlackMath Dec 10 '18 at 9:24
  • $\begingroup$ A similar questions is found here math.stackexchange.com/questions/2631501/… $\endgroup$ – BlackMath Dec 10 '18 at 9:32
  • $\begingroup$ can you show your solution when $z<1$? $\endgroup$ – Atstovas Dec 11 '18 at 17:40
  • $\begingroup$ It will be $$P[Z_1+Z_2+Z_3 \leq z] = \int_{z_3=0}^{z}\int_{z_2=0}^{z-z_3}\int_{z_1=0}^{z-z_2-z_3}dz_1\,dz_2\,dz_3$$ Can you solve this? I suppose it's straightforward. Start from the most inner integral. $\endgroup$ – BlackMath Dec 11 '18 at 23:52

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