Given 8 runners, the time until they reach the finish line is distributed like so $X_1..X_8 \sim Exp(1)$ and they are independent from each other.
Let $T_1$ denote the time of the runner who got to the finish line first.
And $T_2$ denote the time for the runner who got to the finish line second.
I need to find $$E(T_2 – T_1)$$
Because of the memoryless property we know that the time between the second to arrive and the first to arrive is the time for the second to arrive which is: $min\{X_1..X_8/T_1\} $ which is the minimum of 7 Exponintial variables i.e. $$ T_2-T_1 \sim Exp(7) $$
thus $$E(T_2 – T_1) = 1/7$$
however, the expected value is linear, meaning $$E(T_2 – T_1) = E(T_2) – E(T_1) $$
amd $T_2,T_1$ are distributed exponintially as minimums of exponintials, Exp(7) and Exp(8) respectully.
So we get $$E(T_2 – T_1) = E(T_2) – E(T_1)= 1/7 – 1/8 \neq 1/7 $$
What am I missing here?
Best Answer
Using the Theory of Order statistics you get that
$$f_{T_1,T_2}(x,y)=7 e^{-7y}\cdot 8e^{-x}\cdot\mathbb{1}_{[0;\infty)}(x)\cdot\mathbb{1}_{[x;\infty)}(y)$$
Thus setting
$$Z=T_2-T_1$$
you have that
$$F_Z(z)=\int_0^{\infty}8e^{-x}\left[ \int_x^{x+z}7e^{-7y}dy \right]dx=\dots=1-e^{-7z}$$
that is
$$f_Z(z)=7e^{-7z}$$
Or, in other words,
$$T_2-T_1\sim \text{Exp}(7)$$
Using the same theory you get that
$$f_{T_2}(y)=56(1-e^{-y})e^{-7y}$$
with expectation
$$\mathbb{E}[T_2]=\int_0^{\infty}56y(1-e^{-y})e^{-7y}=\frac{15}{56}=\frac{1}{7}+\frac{1}{8}$$