# Conditional Probability – Calculating Probability Conditioned on Inequality

conditional probabilityconditioninginequalityprobabilityprobability-inequalities

Assume that $$A \sim \mathcal{N}(0, 1)$$, $$B \sim \mathcal{N}(0, 1)$$. I am trying to calculate $$P(A \,|\, A < B)$$.

For the sake of this problem, we can assume that $$A \perp B$$, but (for obvious reasons) $$A \not\perp B | A < B$$ (as, in this case, $$A, B$$ are coupled by a common effect).

I know that I can calculate $$P(A < B)$$ on its own as follows:

$$P(A < B) = \int_{b=0}^\infty\int_{a=0}^b P_A(a)P_B(b) dAdB \tag*{(1)}$$

Beyond this, I know that I can rewrite the conditional probability as follows:

\begin{align} P(A \,|\, A < B) = \frac{P(A, A < B)}{P(A < B)} \tag*{(2)} \end{align}

I'm having trouble putting these two pieces together to come up with an expression for $$P(A | A < B)$$, especially since I cannot factorize the numerator in Equation (2) any further because $$A \not\perp A.

Is there a known closed form expression for $$P(A | A? And if so, how is it derived?

Thanks so much in advance for the help!

Edit: this is not a homework problem. ðŸ˜Š

First, we define an additional random variable $$C = B - A$$, the joint distribution $$p(a, b, c)$$ is given by $$p(a, b, c) = \mathcal{N}(0, 1)(a) \ \times \ \mathcal{N}(0, 1)(b) \times \delta(c - (b - a)).$$ Now, we shall focus on the distribution $$p(a, c)$$, which is given by \begin{aligned} p(a, c) &= \int_{-\infty}^{\infty} db \ \mathcal{N}(0, 1)(a) \ \times \ \mathcal{N}(0, 1)(b) \times \delta(c - (b - a))\\ &= \mathcal{N}(0, 1)(a) \int_{-\infty}^{\infty} db \ \mathcal{N}(0, 1)(b) \times \delta(c - (b - a))\\ &= \mathcal{N}(0, 1)(a) \int_{-\infty}^{\infty} db \ \mathcal{N}(0, 1)(b) \times \delta(c + a - b)\\ &= \mathcal{N}(0, 1)(a) \mathcal{N}(0, 1)(c + a). \end{aligned} Because $$A < B$$ is equivalent to $$C >0$$; the probability density $$p(a, a < b) = p(a, c > 0)$$, which is given by \begin{aligned} p(a, c > 0) &= \int_0^{\infty}dc \ \mathcal{N}(0, 1)(a) \mathcal{N}(0, 1)(c + a)\\ &= \mathcal{N}(0, 1)(a)\int_0^{\infty}dc\mathcal{N}(0, 1)(c + a)\\ &= \mathcal{N}(0, 1)(a)\int_a^\infty dx\mathcal{N}(0, 1)(x)\\ &= \mathcal{N}(0, 1)(a)\left[\frac{1}{2}\left(1 - {\rm erf}\left(\frac{a}{\sqrt{2}}\right)\right)\right], \end{aligned} where $${\rm eft}(x)$$ is the error function. Moreover, the distribution $$p(c)$$ would also be a normal distribution with a variance of $$2$$ (see here). Therefore $$p(c > 0) = 1/2$$. The desired distribution $$p(a \vert c > 0)$$ is then given by \begin{aligned} p(a \vert c > 0) &= \frac{p(a, c > 0)}{p(c > 0)}\\ &= \mathcal{N}(0, 1)(a)\left(1 - {\rm erf}\left(\frac{a}{\sqrt{2}}\right)\right). \end{aligned}