Discussion between Math Lover and Mary Star

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Mary Star

10:26

Using the formula of Wikipedia, do we have the following ? @MathLover

$f_{X_1+X_2}=\int_{\mathbb{R}}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}0\cdot f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}0\cdot f_{X_2}(z-t)\, dt\\ =\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-1}^1\frac{1}{2}f_{X_2}(z-t)\, dt\\ =\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt$

Then we have that $f_{X_2}(z-t)=\frac{1}{5}$ if $z-t\in [0,5]$ and otherwise $0$, or not? @MathLover

Math Lover

11:10

@MaryStar no that's not fully correct. As I explained with the diagram, pdf is different for different sub-domains of Z. Where are you taking care of that?

1 hour later…

Mary Star

12:12

So is it also wrong that $f_{X_1+X_2}=\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt$ ? @MathLover

Math Lover

12:29

It is correct but bounds of x is not correct. it is different bounds for different values of z, which is what is in my answer.

@MaryStar for example, for z in (-1, 1), f_Z(z) = \int_{-1}^z 1/2 * f_Y(z-t) dt and f_Y(z-t) = 1/5

Mary Star

12:47

So we write $f_{X_1+X_2}=\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt$ and then we have to consider the intervals for $z$ ? Or is it wrong till that step ? @MathLover

Math Lover

yes consider the intervals of z and write bounds of x accordingly. I just wrote one for -1 \lt z \lt 1 above

Mary Star

Yes,I mean first we write :
$f_{X_1+X_2}=\int_{\mathbb{R}}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}0\cdot f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}0\cdot f_{X_2}(z-t)\, dt\\ =\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-1}^1\frac{1}{2}f_{X_2}(z-t)\, dt\\ =\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt$

and now we consider the intervals for $z$ , right? We have to write this last integral as a sum of integrals as…

Math Lover

yes but when we write pdf, we have to write in intervals of z as pdf is different in different intervals

Mary Star

Do you mean it as follows ?

$f_{X_1+X_2}=\int_{\mathbb{R}}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-\infty}^{-1}0\cdot f_{X_2}(z-t)\, dt+\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt+\int_{1}^{+\infty}0\cdot f_{X_2}(z-t)\, dt\\ =\int_{-1}^1f_{X_1}(t)f_{X_2}(z-t)\, dt\\ =\int_{-1}^1\frac{1}{2}f_{X_2}(z-t)\, dt\\ =\frac{1}{2}\cdot \int_{-1}^1f_{X_2}(z-t)\, dt\\ = \begin{cases} \frac{1}{2}\cdot \int_{-1}^zf_{X_2}(z-t)\, dt& \text{ for } -1<z<1\\ \frac{1}{2}\cdot \int_{-…

1 hour later…

Math Lover

14:06

@MaryStar -1 < z < 1 and 1 < z < 4 are correct. But for 4 < z < 6, it should be \frac{1}{2}\cdot \int_{z-5}^1 f_{X_2}(z-t) dt

finally the answer should be f(z) = (z + 1) / 10 for -1 < z < 1; f(z) = 1 / 5 for 1 < z < 4 and f(z) = (6-z) / 10 for 4 < z < 6

4 hours later…

Mary Star

17:52

We know that $f_{X_2}(y)=\frac{1}{5}$ if $y\in [0,5]$, right?
When we consider for example the case $-1 < z < 1$ how do we get $f(z) = (z + 1) / 10$ ? @MathLover

Math Lover

18:22

@MaryStar int_{-1}^z (1 / 2) \cdot (1/5) dt = (z - (-1)) / 10

5 hours later…

Mary Star

23:01

How do we get that $z$ is in the interval $[-1,6]$ ? Do we get that because $t$ is in $[-1,1]$ and $z-t$ in $[0,5]$ ? @MathLover

Ah we have : $-1\leq t\leq 1$and $0\leq z-t\leq 5 \Rightarrow t\leq z\leq 5+t \Rightarrow -1\leq t\leq z\leq 5+t \leq 6$, right?

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