# Finding conditional expectation: $\mathrm{E}(aX + b | cX + d < eY + g)$

I'm trying to obtain a formula for the following object, without imposing any distributional assumptions:

$$E(aX + b | cX + d < eY + g)$$ Obviously by linearity of $$E$$, $$\ aE(X| \frac{eY + g - b}{a}) + b$$, given c > 0.

I've derived this, is it correct?:

$$= a\int^\infty_{-\infty} \int^\infty_\frac{eY + g - b}{a} X \frac{f_{x,y}}{f_y}dx dy + b$$

If so, why have I found the following similar formula elsewhere - is this when the two variables $$X, Y$$ are independent?

$$a\frac{1}{P(cX + d < eY + g)} \int^\infty_\frac{eY + g - b}{a} X f(x) dx + b$$

• Why not simplify your question first? A little algebra will readily reduce it to $E(\alpha X + \beta Y + \gamma \mid X \lt 0),$ thereby eliminating half of the unknown parameters and hugely simplifying the calculations. But since you impose no distributional assumptions, the main question is too general to be answerable. BTW, all your formulas do make some distributional assumptions: in particular, you appear to assume $(X,Y)$ has a density. – whuber May 6 at 12:24
• Thanks for the response! I'm happy to assume X and Y jointly distributed by some f(x,y). Am I missing something simple, how can I transform it to the form E( | X<0) you mentioned? – tom May 6 at 12:32
• Make an affine transformation of the variables. In particular, let the first variable be $cX - eY + d-g.$ – whuber May 6 at 12:33

Provided $$e^2+c^2\ne 0,$$ the event $$cX + d \lt eY + g$$ defines a half plane. (When $$e^2+c^2=0,$$ this event is either empty or universal, giving trivial answers.)

This suggests we can simplify the problem with a change of coordinates. To this end, define a new random variable

$$U = cX + (-e)Y + (d-g)$$

so that the conditioning event is $$U \lt 0.$$

The expectation is a linear function of $$X,$$ which now needs to be expressed in terms of $$U$$ and another variable. A nice choice (among many) would be a variable orthogonal to $$U,$$ such as

$$V = eX + cY.$$

We may solve for $$aX+b$$ in terms of $$U$$ and $$V,$$ giving

$$aX + b = \frac{a}{c^2+e^2}\left(c(U + g-d) + eV\right)+b = \alpha U + \beta V + \gamma$$

where

$$\alpha = \frac{ac}{c^2+e^2},\quad \beta = \frac{ae}{c^2+e^2},\quad \gamma = \frac{a(g-d)}{c^2+e^2}+b.$$

As a result, the conditional expectation has been re-expressed as

\eqalign{ E(aX + b\mid cX + d \lt eY + g) &= E(\alpha U + \beta V + \gamma \mid U \lt 0) \\ &= \alpha E(U\mid U \lt 0) + \beta E(V \mid U\lt 0) + \gamma }

This is as far as one can go towards simplification in general, but the following remarks may be of some use.

1. To find these expectations you might need to find the joint distribution of $$(U,V).$$ When $$(X,Y)$$ has a density $$f_{X,Y},$$ $$(U,V)$$ will also have a density and can be expressed in terms of $$f_{X,Y}$$ and the Jacobian of the transformation, equal to $$c^2+e^2.$$

2. The only bivariate calculation is the second term, $$E(V\mid U \lt 0).$$ In terms of $$f_{U,V}$$ that's a relatively simple integral conceptually (assuming $$\Pr(U\lt 0)\ne 0$$), given by $$E(V\mid U \lt 0) = \frac{1}{\Pr(U \lt 0)}\int_{-\infty}^0\left(\int_{-\infty}^{\infty} v\,f_{U,V}(u,v)\,\mathrm{d}v\right)\mathrm{d}u.$$ This is recognizable as an integration of conditional expectations over all negative $$u$$--and that can be convenient, because often regression methods provide explicit formulas for conditional expectations. (E.g., in the bivariate Normal case the conditional expectation is a linear function of $$u.$$)

Use $$E(Z|A)=\frac{E(Z1_A)}{P(A)}$$

#Conditional_expectation_with_respect_to_an_event

So without any assumption about distribution of $$(X,Y)$$,By defining $$A=\{ cX + d < eY + g\}$$ we have

$$E(aX + b | cX + d < eY + g)=E(aX + b | A)=a E(X | A)+b=a\frac{E(X 1_A)}{P(A)} +b=a\frac{E(X 1_A)}{P(cX + d < eY + g)} +b$$

$$a\frac{\int_{\{\omega \in \Omega\mid cX(\omega) + d < eY(\omega) + g \}}X \, dP}{P(cX + d < eY + g)} +b$$