Show that conditional expectation is $ P[ E < 7-T-B \mid T,B] = \max(0, 1-e^{-(7-T-B)})$

Question

I am reading the book A Probabilistic Theory of Pattern Recognition and I try to understand his calculation in Section 2.3 Another Simple Example.

In this section, he has three random Variables $T,B,E$ wihch are i.i.d exponential random variables( i.e. they have density $e^{-u}$ on $[0,\infty)$. He claims that a simple calculation shows that

$$ P[ E < 7-T-B \mid T,B] = \max(0, 1-e^{-(7-T-B)})$$

Can someone explain me what exactly happened here?

Answer 1

The property you are interested in follows from the more general fact below, considering $X=E$, $Y=(T,B)$, and $h(t,b)=7-t-b$. One sees that the distribution of $(T,B)$ is irrelevant, as long as $(T,B)$ is independent of $E$ and $E$ is exponentially distributed.

Assume that $X$ and $Y$ are independent random variables and that $X$ has CDF $F_X$, then, for every measurable function $h$, $$P(X<h(Y)\mid Y)=F_X(h(Y))$$

To prove this, recall that, by definition, $P(X<h(Y)\mid Y)=g(Y)$ where the measurable function $g$ is characterized by the fact that, for every $y$, $$P(X<h(Y),Y<y)=E(g(Y);Y<y)$$ Now, by definition of the distribution $P_Y$ of $Y$, the RHS is $$\int_{-\infty}^yg(z)dP_Y(z)$$ and, by independence, the LHS is $$\int_{-\infty}^y\left(\int_{-\infty}^{h(z)}dP_X(x)\right)dP_Y(z)=\int_{-\infty}^yF_X(h(z))dP_Y(z)$$ By identification, this proves that $g=F_X\circ h$, as desired.

Answer 2

$\mathsf P(T+B+E<7|T,B)$ is the probability that $T+B+E<7$ given the values of $T$ and $B$. Thus $T$ and $B$ are constants. Now you can solve the equation for $E$.

$T+B+E<7 \Rightarrow E<7-T-B$

And $E$ is exponentially distributed.

$\mathsf P(E<x)=1-e^{-x}, x>0$

For $x=7-T-B$ we get

$$\mathsf P(E<7-T-B|T,B)=1-e^{-(7-T-B)}=1-e^{T+B-7} \quad \text{on the event}\ \{7-B-T >0\}$$