# Finding a distribution function of random variable sum

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

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

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

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

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

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

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

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

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

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

• How to integrate this? – Atstovas Dec 10 '18 at 9:00
• 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. – BlackMath Dec 10 '18 at 9:24
• A similar questions is found here math.stackexchange.com/questions/2631501/… – BlackMath Dec 10 '18 at 9:32
• can you show your solution when $z<1$? – Atstovas Dec 11 '18 at 17:40
• 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. – BlackMath Dec 11 '18 at 23:52