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

Question

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.

Answer 1

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.

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

Answer 2

Denote $\sum_{i = 1}^{k - 1} X_i$ by $Y$ and $X_k$ by $Z$, by iid assumption and the reproduce property of normal distribution, $Y \sim \mathcal{N}(0,(k - 1)\sigma^2)$. The required probability is \begin{align} P(Y + Z < a | Y < a) = \frac{P(Y + Z < a, Y < a)}{P(Y < a)} \end{align} in which the denominator is easy to evaluate and you can use a double integration to evaluate the numerator (note $Y$ and $Z$ are independent).

In detail, let $\phi$ and $\Phi$ be the pdf of cdf of a standard normal random variable respectively, it follows that \begin{align} & P(Y + Z < a, Y < a) \\ = & \int_{-\infty}^\infty \int_{-\infty}^{a - z} \phi_Y(y)\phi_Z(z)dy dz \\ = & \int_{-\infty}^\infty \Phi\left(\frac{a - z}{\sqrt{k - 1}\sigma}\right) \phi\left(\frac{z}{\sigma}\right) dz % = & \frac{1}{2\pi\sqrt{k - 1}\sigma^2}\int_{-\infty}^\infty \int_{-\infty}^{a - z} \exp\left[-\frac{1}{2(k - 1)\sigma^2}y^2 - \frac{1}{2\sigma^2}z^2\right]dy dz \end{align} the denominator is just $\Phi\left(\dfrac{a}{\sqrt{k - 1}\sigma}\right)$.