Conditional probability of the sum of r.v.

Question

I have $n$ independent random variables $X_i$ with known PDF and CDF (say, Normal, but not necessarily with the same parameters).

Given $U_1, U_2 \subseteq \{1,...,n\} $ such that $U_1 \cup U_2 = \{1,...,n\}$ and $U_1 \cap U_2 = \emptyset $ I need to evaluate the probability $\Gamma$ defined as:

$\begin{equation} \Gamma = P(\sum_{i=1}^n X_i \geq A \;\mid \; X_j\leq \alpha \;\forall j \in U_1, X_k\geq \beta \;\forall k \in U_2 ) \end{equation}$

The only practical approach I can think of is based on Monte Carlo simulation.

My questions are:

1. How can I compute $\Gamma$ analytically, knowing the PDFs/CDFs, for any $n$?
2. Are there other ways (besides Monte Carlo) to approximate $\Gamma$?
3. (Later edit after the first answer) What if $\alpha$ and $\beta$ can be different for each r.v. (i.e. $X_j\leq \alpha_j, X_k \geq \beta_k$)?

Thanks

Answer 1

Assuming $X_i$ are iid, with PDF $f(x)$ and CDF $F(x)$, the conditioning leaves us with $n_1$ and $n_2$ variables with respective PDF:

$$f_1(x) = \frac{1}{F(\alpha)}f(x) \, [x \le \alpha]$$ $$f_2(x) = \frac{1}{1-F(\beta)}f(x) \, [x > \beta]$$

Let $Y=\sum X_i$, the PDF $f_Y$ is formally given by the $n_1$-convolution of $f_1(x)$ with $n_2$ convolution of $f_2(x)$. This can be expressed more compactly with characteristic functions $\psi_Y(w)= \psi_1(w)^{n_1} \psi_2(w)^{n_2}$. But in any case, the calculation can be difficult.

A trivial approximation would be, if $n$ is large, to apply the CLT; this would only require to compute the mean and variance of the truncated PDF $f_1(x)$, $f_2(x)$.