Expected Value and Exponential memoryless property

Question

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?

Answer 1

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 $\int\limits_0^\infty t(56 e^{-7t}-56e^{-8t})\, dt =\frac87 - \frac78 = \frac17+\frac18$

Answer 2

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,

$$T_2-T_1\sim \text{Exp}(7)$$

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Using the same theory you get that

$$f_{T_2}(y)=56(1-e^{-y})e^{-7y}$$

with expectation

$$\mathbb{E}[T_2]=\int_0^{\infty}56y(1-e^{-y})e^{-7y}=\frac{15}{56}=\frac{1}{7}+\frac{1}{8}$$