Why square root of leadtime during safety stock calculation During future demand calculations, the safety stock concept is usually used.
One way to model the safety stock is the following:
SS = Z * sqrt(PC) * std(D)
Where, SS is the safety stock, Z is the Z-score for a given product, PC is the performance cycle (or total lead time), and std(D) is the demand standard deviation.
My question regards why the formula uses sqrt of the Lead Time instead of the pure Lead Time?
Any proof or any book/paper that shows the main idea why to use the sqrt instead of the total leadtime?
 A: As Glen mentioned, additivity of variances when samples are independent: 
VAR(ND) = NVAR(D), 
with VAR = S^2
then S(N*D) = SQRT(N)*S(D)
In statistics, the standard error SE or standard deviation for a sample mean is given by the formula SE = S/SQRT(N) for the same reason as above, where N is the sample size and S is the standard deviation of the sample.
Applying this to supply chain, the standard deviation over a lead time of N days is equivalent as pulling a sample of N days and measuring the standard error of its daily demand:
Average SE for N days = S(D)/SQRT(N), where S(D) is the daily demand standard deviation. Notice this is an average daily standard error. 
The total SE over the length of N days lead time is N times that value = 
SE = N*S(D)/SQRT(N) = SQRT(N)*S(D)
The safety stock is then simply a measure of how many standard deviations away from the mean you want to be = Z*SE (standard error over the lead time).
A: I think your instinct is correct, and the formula should have an L and not a sqrt(L).  I think this stems from the safety stock formula on Wikipedia.  I humbly suggest that it's wrong.   
The mass of depleted quantity over a period L is equal to L*D, where L has units of time and D has units of mass/time.  
The depleted quantity has variability, so we want to protect for the chance it's greater than we expect.  The critical point for the depleted quantity's variability is the safety stock, and this is simply Z*sd(LD).  
The standard deviation of the product of L and D may not be know, and it may be preferable to represent the safety stock with statistics on L and D individually.  In order to do this, must assume that L and D are independent random variables.  
The property of the standard deviation of the product of two random variables is 
sd(XY) = sqrt(E(X)^2 var(Y) + E(Y)^2 var(X) + var(X)var(Y)).  
Therefore the safety stock = 
Z * sqrt(L^2 var(D) + D^2 var(L) + var(D)var(L))
I assume at this point that the assumption is made that the var(D)var(L) term is much smaller than the first two terms, and it is dropped.  This should be the case unless L and D have very high variability.
I think the safety stock should be
Z * sqrt(L^2 var(D) + D^2 var(L))
The formula on Wikipedia is 
Z * sqrt(L var(D) + D^2 var(L))
There's another way to see the error in the Wikipedia formula: by considering the units.  Safety stock has units of mass.  For the Wikipedia formula, the quantity L*var(D) has units of mass^2/time.  This is inconsistent with the units of the D^2*var(L) term, which has units of mass^2.  The first term has to have consistent units with the second term in order to add these terms, and the units have to be mass^2 in order to produce a safety stock with units of mass.  With an L^2 instead of an L, the units work out appropriately.
A: The standard deviation of the sum of Xi from observation 1 to n = sigma * square root of n, so that is why the safety stock is Z * sigma * square root of n, where n = days of lead time.
