The problem said:
If X1,X2,X3 are independent random variables that are uniformly
distributed on (0,1), find the PDF of X1 +X2 +X3.
The theory I have said:
Following the theory and the example for the sum of two random variables, I try to set up the integral, therofore:
$f_{X_1}(a)=f_{X_2}(a)=f_{X_3}(a)= 1$ when $0 < a < 1$ and $0$ otherwise ;
I belive the region of the integral will be a cube of $1 x 1 x 1$, then:
$$P((X_1\leq x_1, X_2\leq x_2, X_3\leq x_3) = \iiint_{\Bbb [0;1]^3} f_{X_1,X_2,X_3}(x_1,x_2,x_3) \;\mathrm d\,x_1 \;\mathrm d\,x_2 \;\mathrm d\,x_3$$
because are independient I have and from the definition of convolution the pdf is:
$$f_{X_1+X_2+X_3}(a)= \int_0^1 \int_0^1 f_{X_1}(a-x_2-x_3) f_{X_2}(x_2) f_{X_3}(a-x_2)\;\mathrm d\,x_2 \;\mathrm d\,x_3$$
Well I am stuck in this part, as I have no idea how to calculate the correct PDF of X1,X2,X3 fx1,x2,x3 from here or if all my procerdure is wrong.Thanks
Best Answer
Because $X_1,X_2,X_3$ are each restricted to values in the interval $(0,1)$, to calculate the pdf of $X_1+X_2+X_3,\; f(a),$ there are different cases depending on $a$ as it ranges over its possible values from $0$ to $3$.
You may like to plot in 3D the plane $x+y+z=a$ for various $a$ values to see why this is and also to determine the limits of integration (though it can be done without plotting).
For $0\leq a\leq 1$:
\begin{align} f(a) &= \int_{x_2=0}^{a} \int_{x_3=0}^{a-x_2} f_1(a-x_2-x_3)f_2(x_2)f_3(x_3)\;dx_3\;dx_2 \\ &= \int_{x_2=0}^{a} \int_{x_3=0}^{a-x_2} 1\;dx_3\;dx_2 \\ & \\ &= a^2/2. \end{align}
The individual pdfs in the integrands are always just $1$ so we use that from here on.
For $1\lt a\leq 2$:
\begin{align} f(a) &= \int_{x_2=0}^{a-1} \int_{x_3=a-1-x_2}^{1} 1\;dx_3\;dx_2\; +\; \int_{x_2=a-1}^{1} \int_{x_3=a}^{a-x_2} 1\;dx_3\;dx_2 \\ & \\ &= -a^2 + 3a - 3/2. \end{align}
For $2\lt a\leq 3$:
\begin{align} f(a) &= \int_{x_2=a-2}^{1} \int_{x_3=a-1-x_2}^{1} 1\;dx_3\;dx_2 \\ & \\ &= a^2/2 - 3a + 9/2. \end{align}