Calculate density using convolution formula

[mathmari]

Hey!

Let $X_1, X_2, X_3$ be i.i.d. with $X_1 \sim U[0, 1]$. I want to determine the density of $S=X_1+X_2+X_3$ using the convolution formula.

I have done the following:

Since $X_1, X_2, X_3$ are i.i.d. we have that they are independent identically distributed random variables. Since $X_1 \sim U[0, 1]$ we have that $X_1$ is distributed uniformly on the interval $[0,1]$.

So, all random variables $X_1, X_2, X_3$ are distributed uniformly on the interval $[0,1]$, right?

Then for the distribution we have that $$f_{X_1}(x)=f_{X_2}(x)=f_{X_3}(x)=\left\{\begin{matrix}
1 & \text{ if } 0\leq x\leq 1\\
0 & \text{ otherwise }
\end{matrix}\right.$$ To determine the density of $S=X_1+X_2+X_3$ we do the following:

We calculate first the density of $Y:=X_1+X_2$ using the convolution formula.

\begin{align*}f_Y(y)&=\int_{-\infty}^{+\infty}f_{X_1}(x)f_{X_2}(y-x)dx \\ & = \int_0^1f_{X_2}(y-x)dx \\ & = \int_0^1 1_{\{0\leq y -x\leq 1\}}dx \\ & = \int_0^1 1_{\{y-1\leq x\leq y\}}dx \end{align*}

At some notes I saw that this integral is equal to $$\left\{\begin{matrix}
y ,& 0\leq y\leq 1\\
2-y ,& 1\leq y\leq 2\\
0 ,& \text{ otherwise }
\end{matrix}\right.$$ How do we get that result? (Wondering) We apply again the convolution formula for $f_Y:=f_{X_1+X+2}$ and $f_{X_3}$.

\begin{align*}f_S(s)&=\int_{-\infty}^{+\infty}f_{Y}(x)f_{X_3}(s-x)dx \\ & = \int_0^1xf_{X_3}(s-x)dx+\int_1^2(2-x)f_{X_3}(s-x)dx \\ & = \int_0^1 x\cdot 1_{\{0\leq s -x\leq 1\}}dx +\int_1^2(2-x)\cdot 1_{\{0\leq s -x\leq 1\}}dx \\ & = \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx +\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx\end{align*}

Is everything correct do far? How can we calculate the last integrals? (Wondering)

 

[I like Serena]

\begin{align*}f_Y(y)&= \int_0^1 1_{\{y-1\leq x\leq y\}}dx \end{align*}

At some notes I saw that this integral is equal to $$\left\{\begin{matrix}
y ,& 0\leq y\leq 1\\
2-y ,& 1\leq y\leq 2\\
0 ,& \text{ otherwise }
\end{matrix}\right.$$ How do we get that result?

Hey mathmari! (Smile)

We have to split the integral into cases:
\begin{align*}f_Y(y)&= \int_0^1 1_{\{y-1\leq x\leq y\}}dx \\
&= \begin{cases}
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }y< 0\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }0 \le y< 1\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if }1 \le y < 2\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if } 2 \le y\\
\end{cases}
\end{align*}
Can we calculate it now? (Wondering)

 

[mathmari]

We have to split the integral into cases:
\begin{align*}f_Y(y)&= \int_0^1 1_{\{y-1\leq x\leq y\}}dx \\
&= \begin{cases}
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }y< 0\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }0 \le y< 1\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if }1 \le y < 2\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if } 2 \le y\\
\end{cases}
\end{align*}
Can we calculate it now? (Wondering)

If $y<0$ then $x$ is not in the interval $[0,1]$, so we have that $\int_0^1 1_{\{y-1\leq x\leq y\}}dx=\int_0^1 0dx=0$.

If $0 \le y< 1$ then we have that $-1\leq y-1\leq x\leq y<1$. Do we have to split it further?

If $1 \le y < 2$ then $0\leq y-1\leq x\leq y<2$. At the part $1<x<2$ the function is $0$. So, do we have to split this also further?

If $2 \le y$ then $x$ is bigger than in the interval $[0,1]$, so we have that $\int_0^1 1_{\{y-1\leq x\leq y\}}dx=\int_0^1 0dx=0$.

(Wondering)

 

[I like Serena]

If $y<0$ then $x$ is not in the interval $[0,1]$, so we have that $\int_0^1 1_{\{y-1\leq x\leq y\}}dx=\int_0^1 0dx=0$.

If $2 \le y$ then $x$ is bigger than in the interval $[0,1]$, so we have that $\int_0^1 1_{\{y-1\leq x\leq y\}}dx=\int_0^1 0dx=0$.

Good! (Nod)

If $0 \le y< 1$ then we have that $-1\leq y-1\leq x\leq y<1$. Do we have to split it further?

If $1 \le y < 2$ then $0\leq y-1\leq x\leq y<2$. At the part $1<x<2$ the function is $0$. So, do we have to split this also further?

No need to split it further.
Just consider that the bounds of our integral are for $0\le x < 1$. (Thinking)

 

[mathmari]

No need to split it further.
Just consider that the bounds of our integral are for $0\le x < 1$. (Thinking)

Ah do we want that $[0,1]\subseteq [y-1, y]$ ? (Wondering)

 

[I like Serena]

Ah do we want that $[0,1]\subseteq [y-1, y]$ ? (Wondering)

It's more like that we want to look at the interval $[0,1]\cap [y-1, y]$ in each case ... (Thinking)

 

[mathmari]

It's more like that we want to look at the interval $[0,1]\cap [y-1, y]$ in each case ... (Thinking)

If $y<0$ then the intersection is empty. The same holds when $y>2$. If $0\leq y\leq 2$ then the intersection is not empty. Right?
Why do we split the interval $[0,2]$ into two intervals?

(Wondering)

 

[I like Serena]

If $y<0$ then the intersection is empty. The same holds when $y>2$. If $0\leq y\leq 2$ then the intersection is not empty. Right?
Why do we split the interval $[0,2]$ into two intervals?

Because the bounds are different. I think we'll see when we work it out. (Thinking)

 

[mathmari]

We have the following:
\begin{align*}&\begin{cases}
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }y< 0\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx & \text{if }0 \le y< 1\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if }1 \le y < 2\\
\int_0^1 1_{\{y-1\leq x\leq y\}}dx& \text{if } 2 \le y\\
\end{cases} \\ & =\begin{cases}
\int_0^1 0dx & \text{if }y< 0\\
\int_0^y dx & \text{if }0 \le y< 1\\
\int_{y-1}^1 dx& \text{if }1 \le y < 2\\
\int_0^1 0dx& \text{if } 2 \le y\\
\end{cases} \\ & =\begin{cases}
0 & \text{if }y< 0\\
y & \text{if }0 \le y< 1\\
1-(y-1)=2-y& \text{if }1 \le y < 2\\
0& \text{if } 2 \le y\\
\end{cases}
\end{align*}

since when $0\leq y<1$ then $[y-1,y]\cap [0,1]=[0,y]$ and when $1\leq y<2$ then $[y-1,y]\cap [0,1]=[y-1,1]$, right? (Wondering)

 

[I like Serena]

since when $0\leq y<1$ then $[y-1,y]\cap [0,1]=[0,y]$ and when $1\leq y<2$ then $[y-1,y]\cap [0,1]=[y-1,1]$, right?

Right! (Nod)

Wait! Doesn't that look like the result we were supposed to get? (Wait)

 

[mathmari]

Right! (Nod)

Wait! Doesn't that look like the result we were supposed to get? (Wait)

Ah ok! Wir haben folgendes:

We apply again the convolution formula for $f_Y:=f_{X_1+X+2}$ and $f_{X_3}$.

\begin{align*}f_S(s)&=\int_{-\infty}^{+\infty}f_{Y}(x)f_{X_3}(s-x)dx \\ & = \int_0^1xf_{X_3}(s-x)dx+\int_1^2(2-x)f_{X_3}(s-x)dx \\ & = \int_0^1 x\cdot 1_{\{0\leq s -x\leq 1\}}dx +\int_1^2(2-x)\cdot 1_{\{0\leq s -x\leq 1\}}dx \\ & = \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx +\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx\end{align*}

Wir bekommen folgdnes :
\begin{align*}f_S(s)&=\int_{-\infty}^{+\infty}f_{Y}(x)f_{X_3}(s-x)dx \\ & = \int_0^1xf_{X_3}(s-x)dx+\int_1^2(2-x)f_{X_3}(s-x)dx \\ & = \int_0^1 x\cdot 1_{\{0\leq s -x\leq 1\}}dx +\int_1^2(2-x)\cdot 1_{\{0\leq s -x\leq 1\}}dx \\ & = \int_0^1 x\cdot 1_{\{-s\leq -x\leq 1-s\}}dx +\int_1^2(2-x)\cdot 1_{\{0\leq s -x\leq 1\}}dx \\ & = \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx +\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx\end{align*}
Wir berechnen die zwei Integral getrennt:

• $\displaystyle{\int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx}$ :
  \begin{align*}\int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx&=\begin{cases}
  \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx & \text{wenn }s< 0\\
  \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx & \text{wenn }0 \le s< 1\\
  \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx& \text{wenn }1 \le s < 2\\
  \int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx& \text{wenn } 2 \le s\\
  \end{cases} \\ & = \begin{cases}
  \int_0^1 x\cdot 0dx & \text{wenn }s< 0\\
  \int_0^s x dx & \text{wenn }0 \le s< 1\\
  \int_{s-1}^1 xdx& \text{wenn }1 \le s < 2\\
  \int_0^1 x\cdot 0dx& \text{wenn } 2 \le s\\
  \end{cases} \\ & = \begin{cases}
  0& \text{wenn }s< 0\\
  \frac{s^2}{2} & \text{wenn }0 \le s< 1\\
  \frac{1}{2}-\frac{(s-1)^2}{2}& \text{wenn }1 \le s < 2\\
  0& \text{wenn } 2 \le s\\
  \end{cases} \\ & = \begin{cases}
  \frac{s^2}{2} & \text{wenn }0 \le s< 1\\
  \frac{1}{2}-\frac{s^2-2s+1}{2}& \text{wenn }1 \le s < 2\\
  0& \text{sonst } \\
  \end{cases} \\ & = \begin{cases}
  \frac{s^2}{2} & \text{wenn }0 \le s< 1\\
  -\frac{s^2-2s}{2}& \text{wenn }1 \le s < 2\\
  0& \text{sonst } \\
  \end{cases}\end{align*}
  
• $\displaystyle{\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx}$ :
  \begin{align*}\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx&=\begin{cases}
  \int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx & \text{wenn }s< 1\\
  \int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx & \text{wenn }1 \le s< 2\\
  \int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx& \text{wenn }2 \le s < 3\\
  \int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx& \text{wenn } 3 \le s\\
  \end{cases} \\ & = \begin{cases}
  \int_1^2(2-x)\cdot 0 dx & \text{wenn }s< 1\\
  \int_1^s(2-x)dx & \text{wenn }1 \le s< 2\\
  \int_{s-1}^2(2-x)dx& \text{wenn }2 \le s < 3\\
  \int_1^2(2-x)\cdot 0 dx& \text{wenn } 3 \le s\\
  \end{cases} \\ & = \begin{cases}
  0& \text{wenn }s< 1\\
  \left (2s-\frac{s^2}{2}\right )-\left (2-\frac{1^2}{2}\right ) & \text{wenn }1 \le s< 2\\
  \left (4-\frac{2^2}{2}\right )-\left (2(s-1)-\frac{(s-1)^2}{2}\right ) & \text{wenn }2 \le s < 3\\
  0& \text{wenn } 3 \le s\\
  \end{cases} \\ & = \begin{cases}
  2s-\frac{s^2}{2}-\frac{3}{2} & \text{wenn }1 \le s< 2\\
  \left (4-2\right )-\left (2s-2-\frac{s^2-2s+1}{2}\right ) & \text{wenn }2 \le s < 3\\
  0& \text{sonst } \\
  \end{cases} \\ & = \begin{cases}
  2s-\frac{s^2}{2}-\frac{3}{2} & \text{wenn }1 \le s< 2\\
  2-2s+2+\frac{s^2}{2}-s+\frac{1}{2} & \text{wenn }2 \le s < 3\\
  0& \text{sonst } \\
  \end{cases} \\ & = \begin{cases}
  -\frac{s^2}{2}+2s-\frac{3}{2} & \text{wenn }1 \le s< 2\\
  \frac{s^2}{2}-3s+\frac{9}{2} & \text{wenn }2 \le s < 3\\
  0& \text{sonst } \\
  \end{cases}\end{align*}

Wir bekommen also folgendes:
\begin{align*}f_S(s)&=\int_0^1 x\cdot 1_{\{s-1\leq x\leq s\}}dx +\int_1^2(2-x)\cdot 1_{\{s-1\leq x\leq s\}}dx \\ & = \begin{cases}
\frac{s^2}{2} & \text{wenn }0 \le s< 1\\
-\frac{s^2-2s}{2}-\frac{s^2}{2}+2s-\frac{3}{2}& \text{wenn }1 \le s < 2\\
\frac{s^2}{2}-3s+\frac{9}{2} & \text{wenn }2 \le s < 3\\
0& \text{sonst } \\
\end{cases} \\ & = \begin{cases}
\frac{s^2}{2} & \text{wenn }0 \le s< 1\\
-s^2+3s-\frac{3}{2}& \text{wenn }1 \le s < 2\\
\frac{s^2}{2}-3s+\frac{9}{2} & \text{wenn }2 \le s < 3\\
0& \text{sonst } \\
\end{cases}\end{align*}

Us everything correct? (Wondering)

 

[I like Serena]

Dein Ergebnis scheint mit Wolfram übereinzustimmen. (Nod)

 

[mathmari]

Dein Ergebnis scheint mit Wolfram übereinzustimmen. (Nod)

Danke! (Giggle) (Mmm)