Joint Distribution given the sum is negative Suppose I have $X_{1},X_{2},\cdots,X_{n}$ IID from a continuous distribution $F$ having density $f$. I have some ideas on how to find the conditional distribution of $(X_{1},X_{2},\cdots,X_{n})$ given $\sum_{i=1}^{n}X_{i}=t$. But I'm stuck on finding the conditional joint distribution when $\sum_{i=1}^{n}X_{i}\geq 0$. For simplicity I was trying to do the proof when $X_{i}\sim N(0,1)$, and I can find the conditional distribution when $\sum_{i=1}^{n}X_{i}=t$ but I cannot progress beyond that. Any help is appreciated.
 A: Let $\Phi(x)$ and $\phi(x)$ be CDF and PDF of $N(0,1)$ respectively. Now you have
$$
W:=\sum_{i=1}^n X_i \sim N(0,n).
$$
Now,
$$
F(x):=\mathbb{P}(W \leq x|W\geq 0) = \frac{\mathbb{P}(W \leq x,W\geq 0)}{\mathbb{P}(W\geq 0)} = 2 \Phi(x)
$$
for $x\geq 0$, and $F_W(x)=0$ for $x<0$. Now differentiate $F$ and you get that the density of $W|W\geq 0$ is
$$
f(x)= 2\phi(x)1(x\geq 0).
$$
Let $(x_1,\ldots,x_n)$ be such that $\sum_{i=1}^nx_i>0$, then for some $r_i>0$, $i=1,2,\ldots,n$, such that
$$
\mathbb{P}(X_1 \in (x_1-r_1, x_1+r_1),\ldots,X_n\in (x_n-r_n, x_n+r_n),W> 0) = \mathbb{P}(X_1 \in (x_1-r_1, x_1+r_1),\ldots,X_n\in (x_n-r_n, x_n+r_n). 
$$
The equality holds since $\{X_1 \in (x_1-r_1, x_1+r_1),\ldots,X_n\in (x_n-r_n, x_n+r_n)\} \subseteq \{W>0\}$, for small enough $r_i$'s. Now you can use independence and write
$$
\mathbb{P}(X_1 \in (x_1-r_1, x_1+r_1),\ldots,X_n\in (x_n-r_n, x_n+r_n)|W> 0) = \\2\times\prod_{i=1}^n\mathbb{P}(X_i \in (x_i-r_i, x_i+r_i)). 
$$
Can you now complete it from here?
A: For simplicity, put $n=3$, $X_1=X,X_2=Y,X_3=Z$. Notice $(X,Y,Z)\sim f_{XYZ}$ where $$f_{XYZ}(x,y,z)=f(x)f(y)f(z)$$ The joint distribution of $(X,Y,Z)$ conditioned on the event $\{X+Y+Z>0\}$ is easily seen to equal
$$(x,y,z)\longrightarrow \frac{f_{XYZ}(x,y,z)\cdot 1_{\{x+y+z>0\}}}{\iiint_{\{x+y+z>0\}}f_{XYZ}\mathrm{d}V}$$ which is equivalent to
$$(x,y,z)\longrightarrow \frac{f_{XYZ}(x,y,z)\cdot 1_{\{x+y+z>0\}}}{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-x-y}^{\infty}f_{XYZ}(x,y,z)\mathrm{d}z\mathrm{d}y\mathrm{d}x}$$ On the other hand, we get with surface integration that the conditional density of $(X,Y,Z)$ conditioned on the event that $X+Y+Z=t$ is $$(x,y,z) \longrightarrow \frac{f_{XYZ}(x,y,z) \cdot 1_{\{x+y+z=t\}}}{\iint_{\{x+y+z=t\}}f_{XYZ}\mathrm{d}S}$$ which is equivalent to $$(x,y,z) \longrightarrow \frac{f_{XYZ}(x,y,z) \cdot 1_{\{x+y+z=t\}}}{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\sqrt{3}f(x,y,t-x-y)\mathrm{d}x\mathrm{d}y}$$ If you want to use the aforementioned density to evaluate a probability like $$P\Big((X,Y,Z)\in E \big|X+Y+Z=t\Big)$$ you would say $$P\Big((X,Y,Z)\in E \big| X+Y+Z=t\Big)=\frac{\iint_{E \cap \{x+y+z=t\}}f_{XYZ}\mathrm{d}S}{\iint_{\{x+y+z=t\}}f_{XYZ}\mathrm{d}S}$$ Can you generalize this to arbitrary $n$?
