Supose that we have two random variables $X \sim \chi^2(n)$ and $Y \sim \chi^2(k)$.
Let Z = X + Y.
Without using Moment Generating Functions show, that $Z \sim \chi^2(n+k)$
Hint: Use two-dimensional density $f_{X,Y}(x,y)$. Make a transition form two-dimensional variable (X,Y) to (Z,Q). Find edge density $f_Z(z)$.
Please, help me. I have no idea how to solve this :(
What did you try? I can not give you all details but I can give you some hints.
Step 1: You can write joint distribution $f_{X,Y}(x,y)$. I think there is an assumption that $X$ and $Y$ are independent. Then $f_{X,Y}(x,y) = f_{X}(x) f_{Y}(y)$, where $f_{X}(x)$ and $f_{Y}(y)$ are probability density functions of $X$ and $Y$ respectively. Remark: The independent assumption is important, if you don't have this assumption, there is no way to compute $f_{X,Y}(x,y)$.
Step 2: Using $f_{X,Y}(x,y)$ to get the joint distribution of $f_{Z,Q}(z,q)$, where $Z=X+Y$ and you are free to choose $Q$. But to make the problem simple, we choose $Q=X$ or $Q=Y$ in general. From the joint distribution of $(X,Y)$ to the joint distribution of $(Z,Q)$, this is called transformation of random variables and you can find details in any probability book.
Step 3: Once you get the joint distribution of $(Z,Q)$, that is $f_{Z,Q}(z,q)$, the probability density $f_{Z}(z)$ is called the marginal distribution of $f_{Z,Q}(z,q)$. To find $f_{Z}(z)$, we simply sum over all values of $Q$ (using integral in this continuous case). But this still can be found in any probability book.