I've been struggling finding the loss functions, $L(\theta,d_1)$ and $L(\theta,d_2)$, for the following question:

Items I manufacture are either independently flawed with probability, $p$, or perfect. If I send a flawed product to a customer, my loss is \$A. I can choose to either send the item without checking it, denoted as decision, $d_1$, or inspect it with a foolproof method and keep replacing the item/checking it until I find a good one to send, decision, $d_2$. The cost of making an item is \$B and the cost of checking it is \$C. Let $\theta_1$ mean the first item is perfect, $\theta_2$ mean the second item is perfect but the first is flawed, $\theta_3$ mean the third item is perfect but the first two are flawed, etc.

If there were a finite number of states of nature, $\Theta$, I could solve this. I'm guessing I have to find the limit of this infinite series? I'm not sure how to do that though. All tips/help appreciated.

My loss matrix: \begin{table}[] \centering \caption{My caption} \label{my-label} \begin{tabular}{lllll} \cline{3-3} & \multicolumn{1}{l|}{d\_1} & \multicolumn{1}{l|}{d\_2} & & \\ \cline{3-3} \theta\_1 & B & B+C & & \\ \theta\_2 & A+2B & 2(B+C) & & \\ \theta\_r & A+rB & r(B+C) & & \end{tabular} \end{table}


You describe a setup in which there are just two states of nature: "flawed" and "perfect." You will make a binary decision whether to check the item (and react accordingly) or not. We need to work out the costs.

There are four possibilities. Initially, you pay $B$ to manufacture either item. At that point you make the decision:

  1. You decide not to check the item, but send it along to the customer.

    a. If it is perfect, no extra cost is incurred. The total cost is $B$.

    b. Otherwise, an extra cost of $A$ is incurred, for a total of $B+A$.

  2. You decide to check the item at a cost of $C$.

    a. If it is perfect, it will pass the check and the customer will be satisfied. The total cost is $B+C$.

    b. If it is flawed, you reject the item and start the procedure over with a new one. The total cost is $B+C$ plus the expected cost of checking items until one passes.

Let the expected cost of (2b) be $X$. It has two components:

  • A cost of $B+C$ incurred with probability $1-p$ (scenario 2a).

  • A cost of $B+C+X$ incurred with probability $p$ (scenario 2b).


$$X = (B+C)(1-p) + (B+C+X)p.$$

The solution is

$$X = \frac{B+C}{1-p}.$$

We can now work out the costs of each decision. For electing not to check the item, the expected cost in (1) is $$\text{Cost}(\text{no check}) = B(1-p) + (B+A)p = B + Ap.$$ For electing to check the item, the expected cost in (2) is $$\text{Cost}(\text{check}) = X = \frac{B+C}{1-p}.$$ You may take these costs as the loss function or, if you wish, you may express the costs relative to the lowest possible one. That will depend on which of these two costs happens to be smaller (and that depends in a somewhat complicated way on the values of $A$, $B$, $C$, and $p$).

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  • $\begingroup$ This was very, very helpful. I get it now. Thank you very much, @whuber. $\endgroup$ – user2205916 Sep 11 '15 at 20:17

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