Background

EMA requires calculation of the expected moments for censored flow values. For a B17C analysis, this includes the flow intervals and perception thresholds.

A perception threshold is merely a convenient way to define a set of identical flow intervals over multiple years of the analysis. Instead of entering the same flow interval multiple times for each individual year, the set of flow intervals can be entered one time as a perception threshold. A left censored perception threshold corresponds to a flow interval with a low flow value of zero (or tau when skew is positive) and a high flow value equal to the perception threshold. This format allows a perception threshold to be input once rather than entering the same flow interval multiple times.

When a flow value is exact, the values of X, (X-M)2, and (X-M)3 needed to calculate the mean, standard deviation, and skew are known values. When a flow value is censored (or uncertain), the values of X, (X-M)2, and (X-M)3 are uncertain values. EMA uses the expected (or average) values for these uncertain values written as E[X], E[(X-M)2], and E[(X-M)3] where the notation E[ ] means expected (or average) value.

Expected moments are a conditional expectation which means that the average value is calculated given a specified set of conditions. The expected moments are calculated given the low and high values that define the flow interval and the LP3 parameter estimates that define the frequency curve. The expression below can be translated as the expected (or average) value of X given X is in the flow interval (i.e. X is between XL and XH) and given X comes from a frequency curve with parameter estimates of μ, σ, γ. The symbol | means conditional (or given).    

$\underline{E}\left[X \mid X_{L}<X \leq X_{H} ; \mu, \sigma, Y\right]$

Expected moments are basically a weighted average. An expected moment is an estimate of the nth moment averaged over all the possible flow values within the flow interval given the LP3 parameter estimates.

The general equation for a weighted average is shown below.

$\bar{y}=\frac{\sum_{i=1}^{p} w_{i} y_{i}}{\sum_{i=1}^{p} w_{i}}$

https://en.wikipedia.org/wiki/Weighted_arithmetic_mean

An arithmetic average is a special case of a weighted average where the weights have the same value (wi = 1/n). Substituting wi=1/n into the weighted average equation and simplifying gives the familiar equation for an arithmetic average.

$\bar{y}=\frac{\sum y_{i}}{n}$

https://en.wikipedia.org/wiki/Arithmetic_mean

Exercise

Question 7. Calculate the expected value E[x] for the roll of a fair six-sided die using the arithmetic average formula.  

Number Rolled (X)

1

2

3

4

5

6

E[X] = ∑X / n =  3.5

Question 8. Calculate the expected value E[x|Loaded] for the roll of a loaded six-sided die using the weighted average equation. The die has been rigged so that the weights are not equal.  Rolling a 6 is 3x more likely.

Number Rolled (X)

w

w*x

1

1/8

0.125

2

1/8

0.250

3

1/8

0.375

4

1/8

0.500

5

1/8

0.625

6

3/8

2.250

∑w = 1.0, ∑w * x = 4.125, E[X | Loaded] = ∑wx / ∑w = 4.125

Question 9. Calculate the conditional expected value E[X | Loaded; X ≥ 4] for a roll of the loaded six-sided die given the result of the roll is greater than or equal to 4. Use the weighted average equation.

Result (X)

w

w*x

1



2



3



4

1/8

0.500

5

1/8

0.625

6

3/8

2.250

∑w = 0.625, ∑w * x = 3.375, E[X | Loaded; X ≥ 4] = ∑wx / ∑w = 5.400