We have $~X \sim U(-1, 1)$ and $Y \sim U(0, 5)$.
For distribution of $Z = X + Y, ~F_Z(z) = P(Y \lt z - x)$
Now as we must have $0 \lt y \lt z - x \lt 5$,
$(i)$ for $- 1 \lt z \lt 1, - 1 \lt x \lt z $
$(ii)$ For $1 \lt z \lt 4, -1 \lt x \lt 1$
$(iii)$ For $4 \lt z \lt 6$,
if $~-1 \lt x \lt z-5, 0...
Why do we have a double integral if we have just the random variable $Y$ at the left side ? That part is not clear for me. Could you explain that further to me ?
You don't have to write double integral. Knowing $F_Y(z-x) = \frac{z-x}{5}$, you could write as $ \displaystyle \int_{-1}^z \frac{z-x}{5} \cdot \frac 12 ~ dx$
Why does it hold that $\displaystyle F_Z(z) = \int_{-\infty}^{\infty} F_Y(z-x) f_X(x) ~ dx$ ? We have that $F_Z(z)=F_{X+Y}(x+y)=F_{X+Y}(x+(z-x))$, right ? How do we continue to get the above formula ?
That's why we call it convolution - it is operation on two functions that gives a third function. Essentially F_Z(z) = P(Z < z) = P(g(X, Y) < z) and here g(X, Y) = X + Y. Now we write this as P(X + Y < z) = P(Y < z - x)...
Discussion between Math Lover and Mar…
Discussion between Math Lover and Mar…