Given two sum process:
$S_k = X_1 + \ldots + X_k$
$S_{k +1} = X_1 + \ldots + X_{k+1}$
where all $X_i$ are iid random variables
I am required to find $f_{S_k, S_{k+1}}(s_k, s_{k+1})$ e.g. their joint PDF
Attempt:
We know that through independent and stationary increment properties of the sum process, we have that the CDF $F_{S_k, S_{k+1}}$ is given by: \begin{align} F_{S_k, S_{k+1}}(s_k, s_{k+1}) & = \Pr[S_{k} \leq s_k, S_{k+1} \leq s_{k+1}]\\ &= \Pr[S_{k} \leq s_k, S_{k + 1} - S_{k} \leq s_{k+1} - s_k]\\ &= \Pr[S_{k} \leq s_k] \Pr[S_{k + 1} - S_{k} \leq s_{k+1} - s_k]\\ &= \Pr[S_{k} \leq s_k] \Pr[S_{1} \leq s_{k+1} - s_k] \\& = F_{S_k}(s_k)F_{S_1}(s_{k+1}-s_k)\end{align}
Then the PDF is going to be the double partial derivative wrt $s_k, s_{k+1}$
\begin{align} \dfrac{\partial^2 F_{S_k, S_{k+1}}(s_k, s_{k+1})}{\partial s_{k+1} \partial s_{k}} & \triangleq f_{S_k, S_{k+1}}(s_k, s_{k+1})\\ & = \dfrac{\partial ^2 F_{S_k}(s_k)F_{S_1}(s_{k+1}-s_k)}{\partial s_{k+1} \partial s_k}\\ & = \dfrac{\partial f_{S_k}(s_k)F_{S_1}(s_{k+1}-s_k) - F_{S_k}(s_k)f_{S_1}(s_{k+1}-s_k)}{\partial s_{k+1} } \\ & = f_{S_k}(s_k)f_{S_1}(s_{k+1}-s_k) - \dfrac{\partial F_{S_k}(s_k) f_{S_1}(s_{k+1}-s_k)}{\partial s_{k+1} }\end{align}
Where the last two steps are computed using the chain rule.
The desired expression is :
$f_{S_k, S_{k+1}}(s_k, s_{k+1}) = f_{S_k}(s_k)f_{S_1}(s_{k+1}-s_k)$
However, observe that I have a bizarre term. $\dfrac{\partial F_{S_k}(s_k) f_{S_1}(s_{k+1}-s_k)}{\partial s_{k+1} }$
(what is the partial derivative of a PDF again?)
Where did I go wrong and how can I fix my derivation?
What you did wrong:
$$S_k\leq s_k, S_{k+1}\leq s_{k+1} \not\Rightarrow S_{k+1}-S_{k} \leq s_{k+1}-s_{k}$$
My solution:
$$f_{S_k,S_{k+1}}(a,b)=f_{S_k,X_{k+1}}(a,b-a)=f_{S_k}(a)f_{X_{k+1}}(b-a)$$
Last inequality follows from independence of $S_k$ and $X_{k+1}$. Finally, use iid of $X_i$’s and $S_1=X_1$.