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if $X_1$ & $X_2$ are uniform random variables then find pdf of $Z= X_1+ X_2$. $$f(x_1,x_2) = 2quadbig[0 leqslant x_1 leqslant x_2 leqslant 1big]$$ I don't know how to set up the limits of integration. Please help,.. i actually need diagram with proper understanding

Just look to the support of the joint function, $(X_1,X_2)in{(x,y): 0leqslant xleqslant yleqslant 1}$. When $Z=X_1+X_2$ this means:

• the pdf for $Z$ is supported over $0$ to $2$, the minimum and maximum values of the sum.


• 
  

  we can substitute $y:=z-x$ to obtain the support for the joint pdf for $(X_1,Z)$ .

  
  
  

  $$begin{split}(X_1,Z)&in{(x,z):0leqslant zleqslant 2 ~land~ 0leqslant xleqslant z-xleqslant 1}\&in {(x,z):0leqslant zleqslant 2~land~0leqslant xleqslant 1 ~land~ 1-zleqslant xleqslant z/2}\&in {(x,z):0leqslant zleqslant 2~land~max(0,1-z)leqslant xleqslantmin(1, z/2)}\&in {left{(x,z):raise{2ex}{[0leqslant z<1~land~0leqslant xleqslant z/2]~\lor~[1leqslant zleqslant 2~land~z-1leqslant xleqslant z/2]}right}} end{split}$$

  
  

We partition the support for $Z$ at $1$ since $max(0,z-1)=begin{cases}0 &:& z<1\z-1&:&1leqslant zend{cases}$

In this instance there is no concern over $min(1,z/2)$ since it is $z/2$ for all $z$ that are supported values of $Z$.

Using the Jaccobian change of variables transformation, the joint pdf we require is:

$$begin{align}f_{X_1,Z}(x,z)&=begin{Vmatrix}dfrac{partial[x,z-x]}{partial[x,z]}end{Vmatrix}f_{X_1,X_2}(x,z-x)\[1ex] &= f_{X_1,X_2}(x,z-x)\ &= 2cdotmathbf 1_{[0leqslant z<1~land~0leqslant xleqslant z/2]~lor~[1leqslant zleqslant 2~land~z-1leqslant xleqslant z/2]}end{align}$$

So applying the Law of Total Probability (Density) .

$$begin{align}f_{Z}(z) &=int_Bbb R f_{X_1,Z}(x,z)mathsf d x\[2ex]& =int_Bbb R f_{X_1,X_2}(x,z-x)~mathsf d x\[2ex]&= int_{0}^{z/2}2cdotmathbf 1_{0leq z<1}~mathsf d x+int_{z-1}^{z/2}2cdotmathbf 1_{1leq zleq 2}~mathsf d x end{align}$$