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I'm studying the probabilitized EOQ Model (probabilistic static demand) but got stuck in a small intermediate step concerning a standard deviation. It should be obvious but I seem to be missing the main clue.

  • Assumption: The demand per unit time ($D$) is normally distributed with mean $D$ and standard deviation $\sigma$.
  • This implies that: the demand during lead time ($D_L = LD$) ($L$ denotes the fixed lead time) must also be normal with mean $E(D_L) = E(LD) = L \ \cdot \ E(D) = LD$ (correct) and standard deviation $\sigma(D_L) = \sigma(LD) = \sqrt(L^2) \ \cdot \ \sigma(D) = \sqrt(L^2) \ \cdot \ \sigma = L\sigma$.

However, the answer should be $\sigma(D_L) = \sqrt(L) \ \cdot \ \sigma = \sqrt(L\sigma^2)$.

What goes wrong?

(The rule I use is $\sigma(aX) = |a| \ \cdot \ \sigma(X) = \sqrt(a^2) \ \cdot \ \sigma(X)$).

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    Is $D_L$ the sum of $L$ gaussian distributions or the product of $L$ times a single gaussian distribution ? What you are assuming to be the correct answer is the result of the sum of $L$ normally distributed variables of mean $D$ and variance $\sigma^2$. – 2017-01-22
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    It is given that: distribution one-day demand (demand per unit time) ($D$): $N(\mu,\sigma)$. Now I want to derive (for example): distribution five-day demand (lead time ($L$) = 5) ($5D$): $N(5\mu,\sqrt5\sigma)$, but how do they come up with the standard deviation $\sqrt5\sigma$? – 2017-01-22
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    I'm guessing the distribution of five-day demand is given by $D_5 = \sum_{i = 1} ^{5} N(\mu, \sigma^2) $. The sum of two random variables is definitely not the same thing as two times one of them. Wikipedia has a very large number of proofs for normally distributed variables (https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables). – 2017-01-22
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    I think your idea is correct. Thank you for the hint. – 2017-01-22

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The assumption is that the demands per period are uncorrelated. This means you can sum up the variances. This will give you the standard deviation as the standard deviation per period times $\sqrt(L)$ .