Suppose you are given the following joint probability distribution
$ f(x,z) = (8x^2z + z^2) \text{ where } 1 \leq x \leq 2, 1 \leq z \leq 2 $
and you what to find $P(X + Z > 2.5)$. How do you determine the integral values you need when you compute this? I am trying to solve a problem similar to this, but I'm just not sure what the correct integral values would be. In this example I think x would be from 1.5 to 2 and z would be from 2.5-x to 2, but is this the right idea?
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and you what to find P(X+Z>2.5). How do you determine the integral values you need when you compute this?
You need to integrate f(x,z) over the intersection of the support and the criteria: $$\{(x,z): 1{\leqslant}x{\leqslant}2, 1{\leqslant}z{\leqslant}2, x{+}z{>}2.5\}\\[2ex]\text{also writen as}\\[1ex]\{(x,z): 1{\leqslant}x{\leqslant}2, \max(1,2.5{-}x){<}z{\leqslant}2\}$$ This can be split into the union: $$\{(x,z):1{\leqslant}x{<}1.5,2.5{-}x{<}z{\leqslant}2\}\cup\{(x,z):1.5{\leqslant}x{\leqslant}2, 1{<}z{\leqslant}2\}$$
So thus
$$\begin{align}\mathsf P(X+Z>2.5)&=\int_{1\leqslant x\leqslant 2, \max(1,2.5-x)<z\leqslant 2} f(x,z)~\mathrm d(x,z) \\[1ex]&= \int_{(1\leqslant x<1.5, 2.5-x<z\leqslant 2)\vee(1.5\leqslant x\leqslant 2, 1<z\leqslant 2)} f(x,z)~\mathsf d(x,z)\\[1ex] &=\int_1^{1.5}\int_{2.5-x}^2 f(x,z)~\mathrm dz~\mathrm dx+\int_{1.5}^2\int_1^2f(x,z)~\mathrm dz~\mathrm dx\end{align}$$
You would have to draw it out. Always draw out the graph. Instead of $X,Z$, think of $X,Y$ (easier to visualize)
We have $X+Y > 2.5$ which means $Y>2.5-X$.
We also have that the range is $1\leq x \leq 2, 1\leq y \leq 2$
First task is to draw the region bounded by $x,y$, and then draw your probability function and see where they intersect. In this case it is the yellow region.
In this case it might help to break the integral in two, as I have shown in black.
So your integrals could be:
$$\int_{1}^{2}\int^{2.5-x}_{1}dydx+\int_{1}^{2}\int^{2}_{1.5}dydx$$