Module 6 Quiz: Indicators, Honestly
This module took the lines off the chart and reduced each of them to arithmetic: an indicator is a function of prices you already hold, so what it discards is countable, how late it is falls out of its weight list, the level it is read against can be solved for, and whether it used bars that had not printed can be tested in a minute. Six questions, all arithmetic. The last one hands back the twenty bars that close this module’s own series and asks four instruments what they see in them, which is four answers about the same twenty numbers.
Covers: Lessons 48 to 52, and the last twenty of the sixty closes this course has carried since lesson 34.
Every question below hands you numbers and asks for a number back. Work all 6 with a calculator before you scroll to the answers; each answer shows the arithmetic, so a wrong result tells you which step to go back to rather than only that you were wrong.
The questions
1. One window, three orderings
Ten closes, in this order: 101.0, 102.4, 100.6, 102.9, 101.3, 103.5, 102.1, 104.2, 102.8 and 103.6.
Compute eight things about them. The average. The sample standard deviation, dividing by nine. The band two standard deviations either side of the average. The highest and the lowest. The net change, last less first. Lesson 36’s efficiency ratio, which is the absolute net change divided by the sum of the absolute changes between consecutive closes. And a simple oscillator over the nine changes inside the window: a hundred times the average rise over the average rise plus the average fall.
Then do all eight again on the same ten numbers read backwards, and a third time on the same ten numbers sorted into ascending order.
Ask. Which of the eight change when the ordering changes, and by how much?
2. Four weight lists, priced
Every smoothing indicator is a weighted sum of past bars, whatever the menu calls it. Here are four, all with weights adding to one, all reading eight bars.
The first gives every one of the eight bars a weight of one eighth. The second is an exponential average with a smoothing constant of two ninths, whose weight on the bar k back is two ninths times seven ninths to the power k. The third weights the most recent bar 8/36, the one before it 7/36, and so on down to 1/36 on the oldest. The fourth is sold as low-lag: 0.6 on the most recent bar, 0.6 on the one before it, nothing on the next five, and −0.2 on the eighth.
Two numbers fall out of any weight list, and neither needs a backtest. The average delay is the sum of each weight times how many bars back it sits. The share of the input’s noise that survives, when the errors in the input are independent, is the sum of the squared weights.
Ask. What is the delay and the surviving noise share of each? Which two are the same filter by both measures, what does the fast one pay, and what did the fourth buy its speed with?
3. Solve the oscillator for its own level
The oscillator is a hundred times the average rise over the average rise plus the average fall, taken across the last fourteen changes. Nothing in that sentence is a judgement except the word average.
Then a specific window. Fourteen changes, nine of them rises averaging 0.80 each and five falls averaging 1.10 each.
Ask. What ratio of average rise to average fall does a reading of 65 require, and of 75, and of 25? If every bar in the window moved the same distance, how many of twenty bars would have to close up to read 65? And what does the specific window above read?
4. The record as drawn and the record as knowable
Twenty-four closes: 100.0, 101.4, 100.6, 102.2, 101.0, 100.2, 100.9, 99.7, 100.8, 102.1, 103.4, 102.5, 101.8, 103.0, 104.2, 103.6, 102.9, 103.2, 103.5, 104.4, 105.6, 106.8, 106.1 and 105.4.
A bar is a pivot low if its close is below the closes of the two bars before it and below the closes of the two bars after it, and a pivot high if it is above all four. Buy at each pivot low, sell at the next pivot high.
The drawn version does what a finished chart does: it enters and exits at the pivot bar’s own close. The knowable version enters and exits two bars later, at that bar’s close, which is the first moment the pivot could be identified without consulting bars that had not printed.
Ask. How many bars carry a low marker at some point and how many keep it? What do the round trips total in each version, and how many of them win? And does the difference between the two totals account for itself?
5. How near counts as near
Twenty closes and their five-bar simple average, drawn at the bar each window ends on.
| Bar | Close | Five-bar average |
|---|---|---|
| 5 | 101.5 | 101.04 |
| 6 | 102.7 | 101.58 |
| 7 | 102.0 | 101.86 |
| 8 | 101.4 | 101.76 |
| 9 | 102.3 | 101.98 |
| 10 | 103.6 | 102.40 |
| 11 | 102.9 | 102.44 |
| 12 | 103.2 | 102.68 |
| 13 | 102.4 | 102.88 |
| 14 | 103.8 | 103.18 |
| 15 | 104.5 | 103.36 |
| 16 | 103.9 | 103.56 |
| 17 | 104.2 | 103.76 |
| 18 | 103.1 | 103.90 |
| 19 | 103.7 | 103.88 |
| 20 | 104.9 | 103.96 |
The first four closes, which the average needs and the table does not test, are 100.0, 100.6, 101.9 and 101.2.
An approach is a close within some stated distance of the average at that same bar. It held if the close two bars later is further from the line, on the side the close approached from. Bars 19 and 20 have no close two bars later, so the fourteen bars from 5 to 18 are the whole of the evidence.
Ask. How many approaches are there at a tolerance of 0.2, 0.4, 0.7 and 1.0, and what share of each held? And on the same fourteen bars, what share of bars that closed up or down continued in that direction two bars later?
6. The same twenty bars, asked four times
The closes of bars 41 to 60 of the series this course has carried since lesson 34: 101.7, 102.1, 103.1, 104.4, 104.8, 105.6, 106.6, 106.1, 106.7, 105.9, 106.7, 105.9, 106.9, 106.6, 105.8, 107.1, 106.8, 106.0, 107.0 and 106.5.
Four instruments, all of them functions of exactly these numbers. Lesson 36’s efficiency ratio over all twenty. The ten-bar simple average at bar 50 and at bar 60, and which side of it the close sits on. The oscillator over the last fourteen changes, and how many of those fourteen bars closed up. And the side of the ten-bar average the close takes at every one of bars 50 to 60.
Ask. What does each of the four say, and how many times does the last one change its mind before and after you require two closes on the same side?
The answers
Each one is worked in full. Where a figure comes from a lesson rather than from this page, the lesson is named.
1. One window, three orderings
The average is 1,024.4 ÷ 10 = 102.440. The deviations from it, squared and summed, come to 12.784, and 12.784 ÷ 9 = 1.4204, whose square root is 1.1918. So the band runs from 102.440 − 2.384 = 100.056 to 102.440 + 2.384 = 104.824.
| Statistic | As printed | Reversed | Sorted up |
|---|---|---|---|
| Average | 102.440 | 102.440 | 102.440 |
| Standard deviation | 1.1918 | 1.1918 | 1.1918 |
| Two standard deviations below | 100.056 | 100.056 | 100.056 |
| Two standard deviations above | 104.824 | 104.824 | 104.824 |
| Highest and lowest | 104.2 / 100.6 | 104.2 / 100.6 | 104.2 / 100.6 |
| Net change, last less first | +2.6 | −2.6 | +3.6 |
| Efficiency ratio | 0.173 | 0.173 | 1.000 |
| Oscillator | 58.67 | 41.33 | 100.00 |
Five of the eight are the same number in all three columns, and they are the five every band, channel and envelope is built out of. Sorting ten closes into ascending order is the most violent thing you can do to a window short of replacing the numbers, and the average, the deviation, the band and the two extremes do not notice. They are symmetric functions: all 3,628,800 orderings of these ten closes give each of them the same answer.
The three that move are the three that look at the order. Reversing the window rearranges the same nine steps, so the path stays 15.0 points and only the net flips sign: the ten closes travel 15.0 to arrive 2.6 higher, which is a ratio of 0.173, and read backwards they travel 15.0 to arrive 2.6 lower, which is the same 0.173. Sorting does something else entirely, because a sorted window has no steps back: every one of its nine changes is a rise, the path collapses from 15.0 to 3.6, and the ratio is 1.000 by construction rather than by anything the market did.
And the oscillator has an exact relation worth carrying. Reversing a window swaps every rise for a fall of the same size, so the reading becomes 100 minus itself: 58.67 forwards and 41.33 backwards. Those are not two readings of anything. They are one statement about direction, written twice.
Answer. Five never move. The net change goes +2.6, −2.6, +3.6; the ratio 0.173, 0.173, 1.000; the oscillator 58.67, 41.33, 100.00.
2. Four weight lists, priced
| Filter | Mean lag, bars | Share of input variance surviving |
|---|---|---|
| Simple average, eight bars | 3.5000 | 0.1250 |
| Exponential average, smoothing constant 2/9 | 3.5000 | 0.1250 |
| Linearly weighted, eight bars | 2.3333 | 0.1574 |
| Low-lag, with one negative weight | −0.8000 | 0.7600 |
Take the first two together. The simple average’s delay is the average of 0 through 7, which is 3.5000, and its sum of squares is eight copies of one sixty-fourth, which is 0.1250. For the exponential average the two sums have closed forms: the delay is (1 − a) ÷ a, which at a = 2/9 is (7/9) ÷ (2/9) = 3.5000, and the sum of squares is a ÷ (2 − a) = (2/9) ÷ (16/9) = 0.1250. Identical in both numbers. The claim that the exponential average is the faster of the two is not true in either sense that can be made precise.
The linearly weighted one is genuinely faster. Its delay is (n − 1) ÷ 3 = 2.3333 bars, which is 1.1667 bars of speed, and its sum of squares is 17/108 = 0.1574. That is 25.9 per cent more of the input’s noise arriving at the output. The trade is the whole design: nothing buys delay back for free.
Then the fourth, and the point is not that it is bad but that its price is printed on its own weight list. Its delay is 0 × 0.6 + 1 × 0.6 + 7 × (−0.2) = 0.6 − 1.4 = −0.8000 bars, which is a filter claiming to sit ahead of the data it is made of. It pays 0.36 + 0.36 + 0.04 = 0.7600, six times the simple average’s noise for eight tenths of a bar of lean. A negative weight is what buys a delay below what a positive-weight filter can reach, and a filter with a negative weight amplifies rather than suppresses at some frequencies, which on a chart is the overshoot after a fast move when the line goes further than the price did.
Answer. 3.5000 and 0.1250 for the first two, 2.3333 and 0.1574 for the third, −0.8000 and 0.7600 for the fourth.
3. Solve the oscillator for its own level
Set the reading to R and solve. A hundred times G over G plus L equals R exactly when G ÷ L = R ÷ (100 − R). At 65 that is 65 ÷ 35 = 1.8571. At 75 it is 75 ÷ 25 = 3.0000. At 25 it is 25 ÷ 75 = 0.3333.
| Reading | Average rise over average fall | Up bars in twenty, at equal sizes |
|---|---|---|
| 25 | 0.3333 | 5 |
| 65 | 1.8571 | 13 |
| 75 | 3.0000 | 15 |
Now put a size on the moves. If every rise and every fall in the window is the same distance, the averages reduce to counts and the reading becomes a hundred times the share of bars that closed up. So a reading of 65 is 65 per cent of the bars up, and on twenty bars that is thirteen. Not a spike, not an extreme, not a market that has run too far: thirteen bars up and seven down.
The specific window is the reason the shortcut has to be stated with its condition attached. Nine rises at 0.80 total 7.20 and five falls at 1.10 total 5.50, so the reading is 100 × 7.20 ÷ 12.70 = 56.69. Nine up bars in fourteen is 64.3 per cent of the bars, and the reading comes in nearly eight points below that, because the falls are 37.5 per cent bigger than the rises. Counting bars and reading the oscillator are the same thing only when the bars are the same size, and they never are.
One consequence for the level you were handed. Eleven rises and three falls at those same two sizes reads 100 × 8.80 ÷ 12.10 = 72.73, so on this window it takes eleven up bars in fourteen to print an overbought reading. Which is a description of an ordinary advance, arriving after it has happened.
Answer. 1.8571, 3.0000 and 0.3333; thirteen bars of twenty; and the window reads 56.69.
4. The record as drawn and the record as knowable
Take the flicker first. A bar becomes a candidate as soon as it is below the two closes before it, and a provisional marker appears on it the moment one later bar closes higher. Bars 6, 8, 13 and 17 all reach that state. Bar 6 loses it: its close of 100.2 is undercut by bar 8 at 99.7 before the second confirming bar arrives, so the marker vanishes. The other three survive to become pivot lows at 99.7, 101.8 and 102.9. One marker in four on this series is not there afterwards, and the rule leaves every signal provisional for exactly one bar beyond its first appearance, because it looks two ahead.
The pivot highs are bars 4, 11, 15 and 22, at 102.2, 103.4, 104.2 and 106.8. Pairing each low with the next high gives three round trips.
| Entry bar | Exit bar | As drawn | As knowable |
|---|---|---|---|
| 8 | 11 | +3.70 | −0.30 |
| 13 | 15 | +2.40 | −1.30 |
| 17 | 22 | +3.90 | +1.90 |
As drawn the strategy makes 10.00 points on three trades and wins all three. As knowable it makes 0.30 on the same three trades and wins one. Nothing about the rule changed and nothing about the series changed; the only difference is when the signal was allowed to be known.
Now the reconciliation, which is the part that shows the gap is not noise. A pivot low is a bar the next two closes are above, so the price two bars later is above it necessarily, and the entry displacement cannot have the other sign. Here those three are 2.4, 2.4 and 0.6, totalling 5.40. A pivot high is a bar the next two closes are below, so selling it in hindsight sells above what was available, and those three are 1.6, 1.3 and 1.4, totalling 4.30. Together 9.70, which is the whole difference between 10.00 and 0.30, with nothing left over.
And the tell is not the win rate. Measure how far each trade goes against its entry in the first two bars: on the drawn entries that figure is exactly zero all three times, because a pivot low is the lowest close in its neighbourhood and there is nothing below it to trade to. On the knowable entries one of the three goes 0.60 against you. A record with no adverse excursion anywhere in it has usually not found a way to avoid pain. It has usually chosen its entries from the far side of the outcome.
Answer. Four markers, three survive. +10.00 with three winners as drawn, +0.30 with one as knowable, and the 9.70 gap is exactly the displacement at the six entries and exits.
5. How near counts as near
| An approach means within | Approaches | Continued away | Share |
|---|---|---|---|
| 0.2 points | 1 | 1 | 100% |
| 0.4 points | 4 | 2 | 50% |
| 0.7 points | 10 | 5 | 50% |
| 1.0 points | 11 | 5 | 45% |
| No level at all: every bar with a direction | 14 | 6 | 43% |
The first column is the whole of the difference between the rows. The bars are the same bars, the average is the same average, and the series did not change while the table was being computed. Widen the tolerance from two tenths of a point to one point and the same twenty closes go from one approach to eleven, which is an eleven-fold change in how much evidence you appear to have.
The share column is the part to be careful with. At two tenths the line held every approach it got, which sounds like a perfect record and is one observation. At four tenths and at seven tenths it holds half of them. At one point it holds 45 per cent. The column moves between 100 and 45 without pattern, which is what a column of small samples looks like rather than a finding about the line.
Then the number that decides whether any of it means anything. On the same fourteen bars, a bar that has just closed up closes higher again two bars later, or a bar that has just closed down closes lower, on 6 of the 14, which is 43 per cent, with no line drawn and nothing to touch. That is the price of admission. The widest reading clears it by two points on eleven observations, which is not a gap anybody can act on, and lesson 50 priced telling a 50 per cent hold rate apart from a 43 per cent one at hundreds of approaches in each condition.
Answer. 1, 4, 10 and 11 approaches, holding 100, 50, 50 and 45 per cent, against a base rate of 43.
6. The same twenty bars, asked four times
The ratio first. The twenty closes travel 14.4 points step by step and arrive 4.8 higher than they started, so the ratio is 4.8 ÷ 14.4 = 0.3333 exactly. A third of the walking ended up as progress.
The average next. Bars 41 to 50 sum to 1,047.0, so the ten-bar average at bar 50 is 104.70 and the close of 105.9 sits 1.20 above it. Bars 51 to 60 sum to 1,065.3, so at bar 60 the average is 106.53 and the close of 106.5 sits 0.03 below it. Two readings ten bars apart, and the second one is on the other side of the line by three hundredths of a point.
The oscillator third, and it is the one that catches people. Six of the last fourteen bars closed up and eight closed down, so 43 per cent of the bars went the reader’s way. The reading is 54.29, which is above the middle. The rises total 5.70 and the falls 4.80, and the oscillator weighs the sizes rather than counting the bars, so a minority of larger rises outvotes a majority of smaller falls. The shortcut from the third question holds only when the bars are the same size, and here they are not.
| Instrument | Reading | What it is a function of |
|---|---|---|
| Efficiency ratio, twenty bars | 0.3333 | All twenty closes |
| Ten-bar average at bar 50 | 104.70 | Bars 41 to 50 |
| Ten-bar average at bar 60 | 106.53 | Bars 51 to 60 |
| Oscillator at bar 60 | 54.29 | The last fourteen changes |
Then the sides, which is where the description stops being a signal. Across bars 50 to 60 the close changes which side of its own average it is on five times, so the line declares six sides in eleven bars, with a median run of one bar. Ask for two consecutive closes on the same side before declaring anything and the six become one: above, from bar 51, and nothing else ever qualifies. The cost is exactly one more bar of lateness at every turn, on top of the 4.50 the ten-bar average already carries by the second question’s arithmetic.
Four instruments and one series. They do not disagree about anything, because they cannot: every one of them is a function of the same twenty numbers, and applying a function cannot add what the numbers did not contain. What the four differ in is which part of those twenty numbers each of them throws away.
Answer. 0.3333; 104.70 and 106.53, above then below; 54.29 on six up bars in fourteen; and six declared sides collapsing to one.
What this quiz was testing
Whether you can price a line instead of believing it. Handed a window, you produce the statistics that cannot see its ordering and the ones that can; handed a weight list, you produce a delay and a noise share without touching a backtest; handed a level, you solve it for the balance of bars it actually asks for; handed a record, you separate what was drawn from what was knowable and reconcile the difference to the point; handed a hold rate, you ask what tolerance produced it and what the tape does with no line drawn at all.
Module 7 leaves the chart entirely. Lesson 53 starts from the other side of your own fills and prices what the person taking them is doing: a spread that just breaks even is twice the chance the next order knows something, multiplied by how much it knows, which turns a quoted penny into a statement about how many informed orders the quoter is expecting.
Educational only. Trading involves substantial risk of loss. Not financial advice. Past performance does not guarantee future results.