Expected Value and Exponential memoryless property 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?
 A: 
Is it possible to explain why that is? Intuitively I agree, however, I can't wrap my head around this

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
$$\mathbb{E}[T_2]=\int_0^{\infty}56y(1-e^{-y})e^{-7y}=\frac{15}{56}=\frac{1}{7}+\frac{1}{8}$$
A: You have correctly said $E[T_2 - T_1] = \frac17$ (the answer to the question) due to memorylessness
and also correct to say that $E[T_1] =\frac18$
You were wrong to say $E[T_2]=\frac17$ and should have said  $E[T_2]=\frac17+\frac18$
You could extend this to say $E[T_8] = \frac18+\frac17+\frac16+\frac15+\frac14+\frac13+\frac12+\frac11$
Think of it as $8$ Poisson processes each with a rate of $1$, equivalent to a single Poisson process with a rate of $8$.  Then the first event occurs and that part drops out at an expected time of $\frac18$, so you are left with the equivalent of a Poisson process with a rate of $7$ and the first event of that occurs at an additional expected time of $\frac17$.  So overall the expectation is $\frac18+\frac17$.
Alternatively use calculus:

*

*The probability no runners have finished by time $t$ is $(e^{-t})^8$

*The probability one runner has finished by time $t$ is $8(1-e^{-t})(e^t)^7$

*The probability two or more runners have finished by time $t$ is $1-(e^{-t})^8-8(1-e^{-t})(e^t)^7$

*The density for the finishing time of the second runner is the derivative of that  $56 e^{-7t}-56e^{-8t}$

*The expected time for the second runner to finish is $\int\limits_0^\infty t(56 e^{-7t}-56e^{-8t})\, dt =\frac87 - \frac78 = \frac17+\frac18$
