Tag Archives: elliot wave prediction 2014

S&P 500 Elliott Wave Technical Analysis – 12th December, 2014

My main Elliott wave count had expected upwards movement. This is not what happened. The main hourly wave count was invalidated with downwards movement, which is what the alternate expected.

Continue reading S&P 500 Elliott Wave Technical Analysis – 12th December, 2014

S&P 500 Elliott Wave Technical Analysis – 21st November, 2014

Upwards movement and a slight increase in upwards momentum was expected.

Continue reading S&P 500 Elliott Wave Technical Analysis – 21st November, 2014

S&P 500 Elliott Wave Technical Analysis – 13th November, 2014

A small green doji candlestick for Thursday’s session invalidated the main hourly Elliott wave count, but was allowed for in the alternate.

Continue reading S&P 500 Elliott Wave Technical Analysis – 13th November, 2014

Russell 2000 Elliott Wave Analysis – 1st October, 2014

Click on charts to enlarge.

Russell 2000 monthly 2013

Data available extends back only to the end of 1987. Data begins with the 1987 “crash” which I am labeling a cycle degree second wave correction.

A five wave impulse for a super cycle wave (I) ended in March, 2000.

Ratios of cycle degree waves within super cycle wave (I) cannot be accurately calculated without all data from the start of Russell 2000 in 1984.

Super cycle wave (II) is seen as a complete expanded flat correction, with a shorter than usual C wave (which is not truncated because it does move very slightly below the end of cycle wave a). Cycle wave c is 45.87 points longer than 1.618 the length of cycle wave a, a variation just less than 10% and so acceptable. Super cycle wave (II) is a 50% correction of super cycle wave (I). Super cycle wave (I) lasted longer than 12 years (it begins when Russell 2000 began) and super cycle wave (II) lasted 9 years.

At 1,691 super cycle wave (III) would reach 1.618 the length of super cycle wave (I).

I am changing my wave count for RUT within super cycle wave (III) to see cycle wave III as over and cycle wave IV underway. Cycle wave III is 26.76 short of 1.618 the length of cycle wave I.

Cycle wave IV may not move into cycle wave I price territory below 745.95.

Russell 2000 weekly 2013

Super cycle wave (III) is incomplete. Within it cycle wave III is probably complete.

Cycle wave IV looks like it is subdividing as a flat or combination. Within it primary wave B should move higher and may make a new high above 1,213.55. Cycle wave IV may find support at the lower edge of the channel.

Draw the channel using Elliott’s first technique: draw the first trend line from the highs of cycle waves I to III, then place a parallel copy on the low of cycle wave II.

Cycle wave II was a shallow 39% zigzag lasting 10 weeks. Cycle wave IV is showing alternation in depth and structure. So far it has lasted 13 weeks and it is not quite halfway through.

Cycle wave IV may not move into cycle wave I price territory below 745.95.

Russell 2000 daily 2013

The final fifth wave of primary wave 5 to complete cycle wave III subdivides perfectly as a five wave impulse.

The downwards movement for primary wave A fits best as a zigzag, with an expanded flat for intermediate wave (B) in the middle. This sees the downwards movement labeled minor wave B within intermediate wave (B) as a zigzag which has a better fit than my last analysis. However, it is very difficult to see the upwards wave of minor wave C within intermediate wave (B) as a five because the middle of the third wave does not fit. I suspect something is wrong in my wave count at the beginning of this movement, within minute wave i and minuette wave (i), but I cannot move the time frame for this data down to see the subdivisions on the hourly chart.

Intermediate wave (C) subdivides perfectly as a completed impulse. The final fifth wave may yet move lower. Draw a channel about intermediate wave (C): draw the first trend line from the ends of minor waves 1 to 3, then place a parallel copy on the end of minor wave 2. Only when this small channel is clearly breached by at least one daily candlestick above it and not touching the upper trend line will I have any confidence that primary wave A is over.

If cycle wave IV is subdividing as a flat correction then within it primary wave B must reach up to a minimum 90% length of primary wave A at 1,202.36. If primary wave A moves lower this point must be recalculated.

If cycle wave IV is unfolding as a combination then primary wave X has no minimum upwards point.

For both a flat or combination the upwards wave of B or X may make a new high above the start of primary wave A at 1,213.55. There is no upper invalidation point.

B and X waves can be time consuming. I would expect it to last about 13 weeks or longer.

Russell 2000 Elliott Wave Analysis – 14th August, 2014

By request from a member here is an Elliott wave analysis of Russell 2000 using data from Yahoo Finance.

Click on charts to enlarge.

Russell 2000 monthly 2013

Data available extends back only to the end of 1987. Data begins with the 1987 “crash” which I am labeling a cycle degree second wave correction.

A five wave impulse for a super cycle wave (I) ended in March, 2000.

Ratios of cycle degree waves within super cycle wave (I) cannot be accurately calculated without all data from the start of Russell 2000 in 1984.

Ratios within cycle wave III of super cycle wave (I) are: primary wave 3 has no Fibonacci ratio to primary wave 1, and primary wave 5 is 8.39 points short of 1.618 the length of primary wave 1.

Ratios within primary wave 3 of cycle wave III are: intermediate wave (3) has no Fibonacci ratio to intermediate wave (1), and intermediate wave (5) is just 3.81 points longer than equality with intermediate wave (1).

Ratios within intermediate wave (3) of primary wave 3 are: minor wave 3 has no Fibonacci ratio to minor wave 1, and minor wave 5 is 7.15 short of equality with minor wave 3.

Ratios within primary wave 5 are: there is no Fibonacci ratio between intermediate waves (3) and (1), and intermediate wave (5) is 6.44 longer than 0.618 the length of intermediate wave (3).

Super cycle wave (II) is seen as a complete expanded flat correction, with a shorter than usual C wave (which is not truncated because it does move very slightly below the end of cycle wave a). Cycle wave c is 45.87 points longer than 1.618 the length of cycle wave a, a variation just less than 10% and so acceptable.

Within cycle wave a primary wave C is 8.82 longer than 2.618 the length of primary wave A, and I am seeing primary wave B as a double combination: zigzag – X – flat.

Within cycle wave b there is no adequate Fibonacci ratio between primary waves A and C. Primary wave A is an impulse, and primary wave C fits perfectly as an ending expanding diagonal. I am confident this upwards wave for cycle wave b is a clear three wave structure, and so it is part of a larger correction.

Within cycle wave c there are no Fibonacci ratios between primary waves 1, 3 and 5.

Overall super cycle wave (II) is a 50% correction of super cycle wave (I). Super cycle wave (I) lasted longer than 12 years (it begins when Russell 2000 began) and super cycle wave (II) lasted 9 years.

At 1,691 super cycle wave (III) would reach 1.618 the length of super cycle wave (I).

When super cycle wave (III) is finished then a subsequent correction for super cycle wave (IV) may not move into super cycle wave (I) price territory below 614.16.

The weekly chart focusses on more detail within super cycle wave (III).

Russell 2000 weekly 2013

Super cycle wave (III) is incomplete. Within it cycle wave III is also incomplete. Cycle wave III may only subdivide as an impulse.

Ratios within cycle wave I are: primary wave 3 is 9.55 points longer than equality with primary wave 1, and primary wave 5 is just 3.77 points short of equality with primary wave 1.

Ratios within primary wave 3 are: there is no Fibonacci ratio between intermediate waves (3) and (1), and intermediate wave (5) is just 2.05 points longer than 0.618 the length of intermediate wave (3).

Within cycle wave III primary wave 3 has no adequate Fibonacci ratio to primary wave 1. It is likely that primary wave 5 will exhibit a Fibonacci ratio to either of primary waves 1 or 3. At 1,355 primary wave 5 would reach equality in length with primary wave 1.

Primary wave 1 lasted 52 weeks, primary wave 2 lasted 13 weeks, primary wave 3 lasted 126 weeks, and primary wave 4 lasted 10 weeks. None of these waves exhibit a Fibonacci duration nor do their durations exhibit Fibonacci ratios to each other. I would expect primary wave 5 to last about 52 weeks, and so the target at 1,355 may be met in about another 40 weeks.

Ratios within primary wave 3 are: intermediate wave (3) has no Fibonacci ratio to intermediate wave (1), and intermediate wave (5) is 8.35 points longer than 0.236 the length of intermediate wave (3).

Ratios within intermediate wave (3) are: minor wave 3 has no Fibonacci ratio to minor wave 1, and minor wave 5 has no Fibonacci ratio to either of minor waves 3 or 1.

Ratios within minor wave 3 are: minute wave iii is 3.54 points short of 2.618 the length of minute wave i, and minute wave v has no Fibonacci ratio to either of minute waves iii or i.

Ratios within minute wave iii are: minuette wave (iii) is 8.36 points short of equality in length with minuette wave (i), and minuette wave (v) has no Fibonacci ratio to either of minuette waves (i) or (iii).

Ratios within minuette wave (iii) are: subminuette wave iii is 10.9 points longer than equality with subminuette wave i, and subminuette wave v is just 3.12 points short of 0.618 the length of subminuette wave iii.

If intermediate wave (2) moves any lower it may not move below the start of intermediate wave (1) at 1,082.53.

Russell 2000 daily 2013

The daily chart focusses on the start of primary wave 5 within cycle wave III.

What is most clear about this chart is the structure of intermediate wave (1): a clear five wave impulse with an extended third wave. This indicates the trend is upwards at this stage despite a deep second wave correction which followed it.

Ratios within intermediate wave (1) are: there is no Fibonacci ratio between minor waves 3 and 1, and minor wave 5 has no Fibonacci ratio to either of minor waves 3 or 1.

Ratios within minor wave 3 are: minute wave iii has no Fibonacci ratio to minute wave i, and minute wave v is just 1.46 points longer than 0.618 the length of minute wave iii.

Within intermediate wave (2) minor wave C is 5.59 points short of 0.618 the length of minor wave A. Intermediate wave (2) is a typically very deep 81% zigzag structure. It is extremely likely it is over here. At 1,319 intermediate wave (3) would reach 1.618 the length of intermediate wave (1).

I have drawn a base channel about intermediate waves (1) and (2). On the way upwards to the target for intermediate wave (3) downwards corrections should find support at the lower edge of this channel. This channel should not be breached to the downside. Upwards movement should breach the upper edge of the channel.

From this first analysis of Russell 2000 I can say that it does not exhibit Fibonacci ratios between all three actionary waves of its impulses, but it often exhibits Fibonacci ratio between two of the three actionary waves within an impulse. Much like the S&P 500, but better than the Nasdaq. This means that target calculation would increase in accuracy only towards the end of a movement; it is only at the end of a movement that targets can be calculated at multiple wave degrees, and because of the tendency for not all actionary waves to exhibit Fibonacci ratios targets should be calculated at least at two or three wave degrees. The target at 1,319 would and should change once minor waves 3 and 4 within intermediate wave (3) are complete.

Russell 2000 does not seem to exhibit Fibonacci relationships in the number of days, weeks or months a wave takes. This makes estimation of when targets may be met rather difficult.

Russell 2000 does seem to exhibit reasonably normal looking Elliott wave structures, and some of these structures are exceedingly typical looking (like the impulse of intermediate wave (1) here on the daily chart). This means that predicted directions should be quite accurate, if the wave count is correct.

S&P 500 Elliott Wave Technical Analysis – 31st July, 2014

An increase in downwards momentum was expected for the main Elliott wave count. The target at 1,946 was exceeded.

Continue reading S&P 500 Elliott Wave Technical Analysis – 31st July, 2014

S&P 500 Elliott Wave Technical Analysis – 17th July, 2014

The correction may now be complete. The target for the next wave up remains the same.

Summary: I expect upwards movement from here towards 2,010. This may be met in another 23 trading days.

Click on charts to enlarge.

The aqua blue trend lines are critical. Draw the first trend line from the low of 1,158.66 on 25th November, 2011 to the next swing low at 1,266.74 on 4th June, 2012. Create a parallel copy and place it on the low at 1,560.33 on 24th June, 2013. While price remains above the lower of these two aqua blue trend lines we must assume the trend remains upwards. This is the main reason for the bullish wave count being my main wave count.

Bullish Wave Count.

S&P 500 daily 2014

There are a couple of things about this wave count of which I am confident. I see minor wave 3 within intermediate wave (1) as over at 1,729.86 (19th September, 2013). It has the strongest upwards momentum and is just 0.76 longer than 2.618 the length of minor wave 1. At 455 days duration this is a remarkably close Fibonacci ratio. The subdivisions within it are perfect. If this is correct then minor wave 4 ends at 1,646.47 and this is where minor wave 5 begins.

Minor wave 5 may be only one of two structures: a simple impulse or an ending diagonal. At this stage an ending diagonal looks less likely, and most recent upwards movement is looking like a third wave so a more common impulse is more likely. If minor wave 5 has passed its middle then I would expect to see more divergence between price and MACD develop over coming weeks.

Along the way up towards the final target I would expect to see two more corrections complete: the next for minuette wave (iv) and the last for minute wave iv.

At this stage today I expect now that subminuette wave iv is complete. Once this is confirmed with a new high above 1,983.94 I will move the invalidation point on the daily chart up to the beginning of subminuette wave v. While the end of subminuette wave iv is unconfirmed it may continue, and it may not move into subminuette wave i price territory below 1,884.89.

At 2,010 minuette wave (iii) would reach 1.618 the length of minuette wave (i). Within minuette wave (iii) subminuette wave iii is 8.31 points short of 1.618 the length of subminuette wave i, and so I would not expect to see a Fibonacci ratio for subminuettte wave v. I will leave the target calculation at minuette wave degree.

At 2,218 minor wave 5 would reach equality in length with minor wave 3. This target may be met in October.

I have drawn a parallel channel about minuette wave (iii) using Elliott’s first technique: draw the first trend line from the highs of subminuette waves i to iii, then place a parallel copy on the low of subminuette wave ii. Subminuette wave iv may have remained mostly contained within the channel. Subminuette wave v may end about the upper edge of this channel.

Minuette wave (i) lasted 12 days, just one short of a Fibonacci 13. Minuette wave (ii) lasted a Fibonacci 34 days. Minuette wave (iii) so far has lasted 66 days. The next Fibonacci relationship could see it end in a total 89 days, which now is in 23 days time. This would also see minuette waves (i), (ii) and (iii) exhibit Fibonacci ratios to each other, as those ratios are derived by the relationship between Fibonacci numbers.

The large maroon – – – channel is copied over from the weekly chart. It is drawn in exactly the same way on bull and bear wave counts. For the bull wave count this channel is termed a base channel about primary waves 1 and 2. A lower degree second wave should not breach the lower edge of a base channel drawn about a first and second wave one or more degrees higher. The lower maroon – – – trend line differentiates the bull and bear wave counts at cycle degree and monthly chart level.

S&P 500 hourly 2014

At the hourly chart level subminuette wave iv now subdivides nicely as a running contracting triangle (the B-D trend line is sloped and not flat enough to term this a barrier triangle).

Tomorrow micro wave E may move very slightly lower at the start of the session. It may not move beyond the end of micro wave C at 1,952.86 for the triangle to remain valid.

Micro wave E slightly overshoots the A-C trend line which is the second common place for E waves of triangles to end.

If this analysis is correct then we should see a fifth wave begin upwards tomorrow. This would be confirmed with movement above 1,983.94.

This main wave count has a higher probability because it has a great fit and sees subminuette wave iv at 16 days, almost perfectly in proportion to subminuette wave ii at 17 days. They show great alternation: subminuette wave ii was a shallow 32% combination and subminuette wave iv is a more shallow 12% triangle. The orange channel is showing nicely where price is finding support and resistance.

S&P 500 5 minute 2014

Alternate Hourly Wave Count

S&P 500 hourly 2014

I have learned the hard way to not trust triangles on the S&P. Too many times I have labeled a movement as a complete or unfolding triangle only to have the wave count invalidated. I must consider all possibilities.

It may be that only micro wave B within subminuette wave iv is a triangle. Within micro wave B triangle submicro wave (E) may not move beyond the end of submicro wave (C) at 1,983.94.

This alternate sees another day or so of sideways movement, before a five wave movement downwards to complete subminuette wave iv. Subminuette wave iv may end when it slightly overshoots the lower edge of the orange channel copied over here from the daily chart.

At this stage this hourly wave count has a low probability because both the A-C and B-D trend lines of this possible triangle have a shallow slope. Triangle trend lines normally have a steeper slope than this.

Bearish Alternate Wave Count

S&P 500 daily bear 2014

This bearish alternate wave count expects that the correction is not over. The flat correction which ended at 666.79 was only cycle wave a (or w) of a larger super cycle second wave correction.

The structure and subdivisions within primary wave C for the bear wave count are the same as for intermediate wave (1) for the bull wave count. Thus the short to mid term outlook is identical.

The differentiation between the bull and bear wave count is the maroon – – – channel. The bull wave count should see price remain above the lower maroon – – – trend line. The bear wave count requires a clear breach of this trend line. If this trend line is breached by a full weekly candlestick below it and not touching it then this bear wave count would be my main wave count and I would then calculate downwards targets.

We should always assume the trend remains the same until proven otherwise; the trend is your friend. While price remains above the lower maroon – – – trend line I will assume that the S&P 500 remains within a bull market.

This analysis is published about 09:10 p.m. EST.

S&P 500 Elliott Wave Technical Analysis – 13th May, 2014

Upwards movement was expected. The second target is widened to a zone and now has a higher probability than the first target.

Continue reading S&P 500 Elliott Wave Technical Analysis – 13th May, 2014

S&P 500 Elliott Wave Technical Analysis – 28th March, 2014

Some upwards movement was expected to begin Friday’s session. The short term target was at 1,859. Price did move higher for the first two hours, but the target was exceeded with upwards movement ending at 1,866.63. Thereafter, price moved lower for the remainder of the session.

Continue reading S&P 500 Elliott Wave Technical Analysis – 28th March, 2014

S&P 500 Elliott Wave Technical Analysis – 14th March, 2014

Price moved lower as expected, but the short term target was not reached and downwards momentum did not increase.

Continue reading S&P 500 Elliott Wave Technical Analysis – 14th March, 2014