I am looking at a problem form Hogg, Tannis & Zimmerman (Ed. 10), and I am curious if the given problem is calculable (for an upper-level undergrad math/stats course) because of the choice of the parameters.

To respect the authors' copyright, I won't repeat the problem as written, but I will represent the key query: $X_1$ and $X_2$ are i.i.d. with a gamma distribution with parameters shape $k=1$ and scale $\theta=2$ (using the parameterization shown on the Wikipedia page). Let $Y_1=\min(X_1,X_2)$ and $Y_2=\max(X_1,X_2)$ and $Z=\alpha Y_1 + \beta Y_2$. Compute the expected value of $Z$.

My strategy would be to do the following:

  1. Find the cdf $G_2(y)=P(Y_2 \le y) = F(y)^2$
  2. Find the pdf $g_2(y) = G_2^\prime(y)$ (using $f(y) = F^\prime(y)$ and the chain rule)
  3. Find $E(Y_2) = \int_{S_{Y_2}} y g_2(y)\ dy$
  4. Repeat for $E(Y_1)$ using $G_1(y) = (1-F(y))^2$
  5. Use the linearity of the expectation operator to find $E(Z)$.

Because I hoped to solve a more general version of the problem, I left the parameters as $k$ and $\theta$ in the gamma function (instead of using the proposed values). This appears to result in a problem that cannot be easily calculated in terms of these parameters (or the size of the random sample, which I also left as an unspecified value $n$).

However, when you plug these given parameters into the gamma function, you get an exponential function. And, the resulting formula is easily calculated for a sample (at least for size $n=2$).

Again, my query is if a general solution is possible (in terms of the given parameters and closed-form solutions) for specific parameters...or perhaps I am missing something in my approach that would simplify the process.

  • 3
    $\begingroup$ Quoting a textbook problem constitutes fair use; it does not violate any (US) copyright law. $\endgroup$ – whuber Jan 16 at 16:48

I won't demonstrate any more than you already know, but this approach might help confirm your suspicion that the parameter $k=1$ is special. Along the way you might find some useful general shortcuts.

Let's begin far more generally: suppose only that $X_1$ and $X_2$ are independent with distribution functions $F_1$ and $F_2,$ respectively (which means for all real $x,$ $\Pr(X_i\le x) = F_i(x)$). I will assume the expectations of the $X_i$ are known.

You can exploit $Y_1+Y_2 = X_1+X_2$ to express

$$E[Y_1] + E[Y_2] = E[X_1] + E[X_2]$$

and from this compute

$$\begin{aligned} E[\alpha Y_1 + \beta Y_2] &= \alpha(E[Y_1] + E[Y_2]) + (\beta-\alpha)E[Y_2] \\&= \alpha(E[X_1]+E[X_2]) + (\beta-\alpha)E[Y_2]. \end{aligned}$$

This reduces the problem to finding $E[Y_2].$

The event $Y_2\le x$ is the event that both $X_i$ are less than or equal to $x,$ whence

$$F_{Y_2} = \Pr(Y_2\le x) = \Pr(X_1\le x)\Pr(X_2\le x) = F_1(x) F_2(x)$$

due to the independence of the $X_i.$


$$\begin{aligned} E[Y_2] &= \int_{-\infty}^0 F_{Y_2}(y)\mathrm{d}y + \int_0^\infty 1-F_{Y_2}(y)\mathrm{d}y\\ &= \int_{-\infty}^0 F_1(y)F_2(y)\,\mathrm{d}y + \int_0^\infty 1-F_1(y)F_2(y)\,\mathrm{d}y. \end{aligned}$$

When at least one of the $X_i$ has nonnegative support (as in this question), the first integral is zero. Thus, whatever the answer might be is tantamount to evaluating the second integral. That's hard to do when the $F_i$ are incomplete Gamma functions--I don't know of any theory of integrals of products of those functions--but there is an easy special case: when they are exponentials (incomplete Gamma functions with parameter $1$).

Let's deal with that. Let the scale factors be $\theta_i,$ so that $F_i(y) = 1 - \exp(y/\theta_i)$ for all $y\ge 0.$ Plugging into the foregoing expectation formula gives

$$\begin{aligned} E[Y_2] &= \int_{0}^\infty 1 - (1 - e^{y/\theta_1})(1 - e^{y/\theta_2})\,\mathrm{d}y = \int_0^\infty e^{y/\theta_1} + e^{y/\theta_2} - e^{y/\theta_1 + y/\theta_2}\,\mathrm{d}y\\ &= \theta_1 + \theta_2 - \frac{\theta_1\theta_2}{\theta_1+\theta_2}. \end{aligned}$$

Since $E[X_i] = \theta_i,$ the answer is

$$E[\alpha Y_1 + \beta Y_2] = \beta(\theta_1+\theta_2) - (\beta-\alpha)\left(\frac{\theta_1\theta_2}{\theta_1+\theta_2}\right).$$

There are other special cases you might enjoy exploring: for positive integral shape parameters $k,$ repeated integration by parts ($k-1$ times) establishes that the incomplete Gamma functions $F_i$ are polynomials times an exponential. This makes the integrand $1-F_1(y)F_2(y)$ a polynomial (usually of higher degree) times exponentials, which is elementary to integrate. There will be many terms, though: when the shape parameters are $k_i,$ expect around $(1+k_1)(1+k_2)$ nonzero terms.

  • $\begingroup$ This approach with $Y_1+Y_2 = X_1+X_2$ is great (hadn't considered that nice trick)...but it does seem to limit a generalization of this problem to random samples of size $n=2$. Thnx for confirming that this problem does have some very strong reliance on the parameter choice for the gamma distribution. $\endgroup$ – Gregg H Jan 17 at 18:23
  • $\begingroup$ The (useful) generalization is that the expectation of the sum of all the order statistics equals the sum of the expectations. $\endgroup$ – whuber Jan 17 at 18:39
  • $\begingroup$ My thought is that if the problem retains the r.v. $Z$ defined as the sum of the min and max, then the full ordering won't help there. $\endgroup$ – Gregg H Jan 17 at 19:00

You are correct that $\mathsf{Gamma}(\mathrm{shape}=1,\mathrm{scale}=\theta)$ is exponential, specifically $\mathsf{Expm}(\mathrm{mean}=\theta)\equiv\mathsf{Exp}(\mathrm{rate}=1/\theta).$

Using the CDF method, you can show that if $X_1, X_2$ are IID $\mathsf{Exp}(1/\theta),$ then $Y_1=\min(X_1,X_1) \sim \mathsf{Exp}(2/\theta),$ so that $E(Y_1) =\mu_1 =\theta/2.$

Similarly but with a little more difficulty, you can find $E(Y_2) = E(\max(X_1,X_2)) = \mu_2$ (in terms of $\theta).$

Finally, $E(\alpha Y_1 + \beta Y_2) = \alpha\mu_1 + \beta\mu_2.$

Here is a brief simulation with $\theta = 6, \alpha=5, \beta=10$ with sufficiently accurate results (2 or 3 significant digits) that you can check each part as you go along.

th = 6;  alp = 5;  bet = 10
x1 = rexp(10^7, 1/6)
x2 = rexp(10^7, 1/6)
mean(x1);  mean(x2)
[1] 5.999339   # aprx E(X1) = 6
[1] 6.000077   # aprx E(X2) = 6

y1 = pmin(x1, x2)
y2 = pmax(x1, x2)
mean(y1);  mean(y2)
[1] 2.999401   # aprx E(Y1) = 3
[1] 9.000015   # aprx E(Y2) = 3 + 6 = 9

z = alp*y1 + bet*y2
[1] 104.9972   # aprx E(Z) = 15 + 90 = 105

Note: An intuitive answer to $E(Y_2)$ is that that the min averages 3, then by the no-memory property of exponential random variables, the max must be 6 greater on average, so $E(Y_2) = 3+6=9.$

  • $\begingroup$ Thanks for sharing this response. I think you mean in the note that the max averages 3. First, I'm uncertain how the no-memory applies here. Also, it appears you've provided a general solution for $k=1$, but this doesn't address my initial query for a general solution w.r.t. the given parameters. (This seems to confirm my impression that this problem is an artifact of a special set of parameters for the gamma function). $\endgroup$ – Gregg H Jan 17 at 1:53
  • $\begingroup$ The min averages 3. Consider modeling the lifetime of a type of device. Each device is subject to potentially fatal events at a rate of $1/6$ per year or one every 6 years. Then a series system of two devices is hit a rate of $2/6=1/3$ per year or one every three years. So the avg of minimum is 3 yrs. By the no memory property the surviving device has an avg life of another 6 yrs so the max will avg 9 yrs. // Because you haven't shown much of a start on a solution I hoped my Ans would give you a start. // I had NO plan to work the entire textbook problem for you. // Sorry if this not helpful. $\endgroup$ – BruceET Jan 17 at 3:55
  • $\begingroup$ No sure what is meant by "haven't shown much of a start on a solution" ...as the strategy in the original post does indeed produce the correct answer (for $k=1$ and $\theta=2$). Though, as noted, the query was about generalizing the problem (if possible). Thnx for your response. $\endgroup$ – Gregg H Jan 17 at 18:27

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