[Math] conditional probability of a sum or iid normal random variables given a bound on a subset of them

normal distributionprobability

Let $X_i$ be iid normal random variables with mean 0 and standard deviation $\sigma$. Is there a straightforward formula to compute the conditional probability $\mathbb{P}(\sum_{i=1}^{k}X_i < a\:\vert\: \sum_{i=1}^{k-1}X_i < a)$?

If someone can give me a hint, that would be great. Thanks in advance.

Best Answer

Let $S_k = \sum\limits_{i=1}^k X_i$ where $\{X_i\}\mathop{\sim}^\textsf{iid}\mathcal{N}(0,1^2)$

Since $\{X_i\}$ are iid, then $X_k$ and $S_{k-1}$ are independent.

$$\begin{align}\mathsf P(S_k< a\mid S_{k-1}<a) & = \dfrac{\mathsf P(S_k<a \cap S_{k-1}<a)}{\mathsf P(S_{k-1}<a)}\\[1ex] & = \dfrac{\int_\Bbb R \mathsf P(S_{k-1}+x <a \cap S_{k-1}<a)\;f_{X_k}(x)\operatorname d x}{P(S_{k-1}<a)} \\[1ex] & = \dfrac{\int_{-\infty}^0 \mathsf P(S_{k-1}<a)\;f_{X_k}(x)\operatorname d x+\int_{0}^{\infty} \mathsf P(S_{k-1}<a-x)\;f_{X_k}(x)\operatorname d x}{\mathsf P(S_{k-1}<a)}\\[1ex] & = \dfrac 1 2 + \dfrac{\int_{0}^{\infty} \mathsf P(S_{n-1}<a-x)\;f_{X_n}(x)\operatorname d x}{\mathsf P(S_{n-1}<a)}\\[1ex] & = \dfrac 1 2 + \dfrac{\int_{0}^{\infty} \Phi_{0,n-1}(a-x)\;\phi_{0,1}(x)\operatorname d x}{\Phi_{0,n-1}(a)}\end{align}$$

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