MHB Calculate density using convolution formula

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The discussion focuses on calculating the density of the sum of three independent and identically distributed uniform random variables, specifically \( S = X_1 + X_2 + X_3 \) where \( X_i \sim U[0, 1] \). The convolution formula is applied to find the density of \( Y = X_1 + X_2 \), leading to a piecewise function for \( f_Y(y) \). The participants explore the integration process, breaking it into cases based on the value of \( y \) to derive the correct density function. Ultimately, they confirm that their results align with established calculations, validating their approach and findings. The discussion emphasizes the importance of careful case analysis in convolution calculations.
mathmari
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Hey! :o

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)
 
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mathmari said:
\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)
 
I like Serena said:
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)
 
mathmari said:
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)

mathmari said:
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)
 
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
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)
 
mathmari said:
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)
 
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)
 
  • #10
mathmari said:
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)
 
  • #11
I like Serena said:
Right! (Nod)

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

Ah ok! Wir haben folgendes:

mathmari said:
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)
 
  • #12
Dein Ergebnis scheint mit Wolfram übereinzustimmen. (Nod)
 
  • #13
I like Serena said:
Dein Ergebnis scheint mit Wolfram übereinzustimmen. (Nod)

Danke! (Giggle) (Mmm)
 
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