S&P 500 Elliott Wave Technical Analysis – 6th November, 2013

Yesterday’s analysis expected downwards movement which is not what happened. Price has moved higher, but without a new high both wave counts remain valid.

Click on the charts below to enlarge.

Main Wave Count.

S&P 500 daily 2013

This wave count has a higher probability than the alternate. Upwards movement over the last 4 1/2 years subdivides best as a zigzag. If something is “off” about the supposed recovery then it must be a B wave because there is plenty that is off in this scenario in terms of social mood.

Price may have found resistance along the upper edge of the big maroon channel from the monthly chart, if the upper trend line is pushed out to encompass all of primary wave A.

At 1,858.03 cycle wave b would reach 138% the length of cycle wave a. This wave count sees a super cycle expanded flat unfolding, and the maximum common length for a B wave within a flat is 138% the length of the A wave. Above this point this wave count would reduce in probability and it would be more likely that a longer term bull market is underway.

Intermediate wave (5) is either incomplete with just minor wave 1 within it completed (as per the first hourly chart below) or, if we move the degree of labeling within it up one degree, intermediate wave (5) and in turn cycle wave b may now be completed.

However, we should always assume the trend remains the same until proven otherwise. I now have four conditions to be satisfied in looking for a trend change. As each condition is satisfied my confidence in this trend change would increase.

At 1,864 intermediate wave (5) would reach equality in length with intermediate wave (1). This is the most common ratio between first and fifth waves so this target has a good probability.

Within intermediate wave (5) no second wave correction may move beyond the start of the first wave. This wave count is invalidated with movement below 1,646.47.

S&P 500 hourly 2013

Upwards movement may have been a continuation of minute wave b as a double combination: flat – X – zigzag. There is almost no room left for upwards movement now, and if this upwards movement is a correction then it must be over.

At 1,754 minute wave c would reach equality in length with minute wave a.

Minor wave 2 may not move beyond the start of minor wave 1. This wave count is invalidated with movement below 1,646.47.

If this wave count is invalidated with downwards movement then the hourly alternate chart is the wave count I would use.

S&P 500 hourly alternate 2013

By simply moving the degree of labeling within intermediate wave (5) up one degree it is possible to see cycle wave b as complete, and a massive trend change may have occurred just a few days ago.

Within the new downward trend I am labeling the subdivisions on this hourly chart in the same was as the first hourly chart above, because 1-2-3 and A-B-C subdivide in exactly the same way.

For this alternate idea upwards movement is a second wave correction which must be over. This wave count expects an increase in downward momentum which must happen tomorrow. At 1,742 minuette wave (iii) would reach 1.618 the length of minuette wave (i).

In looking for confirmation and confidence in this wave count I will look for the following (in order):

1. Movement below 1,646.47.
2. A clear breach of the black channel on the daily chart containing primary wave C.
3. A clear breach of the larger maroon channel on the monthly chart containing cycle wave b (this would provide enough confidence for me to discard the bullish alternate).
4. Final price confirmation with movement below 1,074.77.

Alternate Bullish Wave Count.

S&P 500 daily alternate 2013

It is possible that we are and have been in a new bull market for a cycle degree fifth wave. Cycle waves should last from one to several years (as a rough guideline).

For this alternate downwards movement may have now begun for intermediate wave (2). This downwards movement must subdivide as a corrective structure, most likely a zigzag.

Intermediate wave (2) may not move beyond the start of intermediate wave (1). This wave count is invalidated with movement below 1,074.77.

If intermediate wave (2) breaches the parallel channel on the monthly chart then a lower degree second wave would breach an acceleration channel about a first wave one degree higher. This would give the wave count a strange look and reduce it’s probability significantly. At that stage I would probably discard it.

4 thoughts on “S&P 500 Elliott Wave Technical Analysis – 6th November, 2013

  1. Hi Lara,
    On last night’s (11-6) hourly alternate chart you swapped out the ABC wave subdivision used in the bullish primary count and re-identified them as minuette waves (i) (ii) (iii) in the bearish alternate count because A-B-C and 1-2-3 subdivide in exactly the same way. Here’s my confusion/question: It’s widely held that a B wave can extend beyond the start of the A wave, by as much as the 1.382 extension, right? At the same time, isn’t it also widely held that a 2 wave cannot extend below the start of the 1 wave? In your alternate hourly chart a B wave comprising a part of the (ii) wave extends below the start of the 1 wave. Is that acceptable in W-X-Y scenarios?

    1. Within the bullish first hourly chart because it is a zigzag (because the a wave subdivides as a five, not a three) then the b wave may not move beyond the START of the a wave. That price point is at 1,775.22.

      Within the bearish second hourly chart the correction is seen as a second wave, and that may not move beyond the START of the first wave. The price point is the same, 1,775.22, and the rule for second waves and B waves (within zigzags) is the same.

      Within the B wave of the first hourly chart there is a price point beyond the END of the a wave below 1,755.72. That is okay.

      Within the second wave of the bearish hourly chart there is a B wave extending beyond the start of the A wave because that structure is a flat; the A wave subdivides as a three. This is not only acceptable but very common within W-X-Y combinations because one of the structures would most likely be a flat, and the most common type of flat is an expanded flat which requires the B wave to be minimum of 105% of the A wave, so it must extend beyond the start of the A wave.

      Does this make sense?

      I’ll explain as clearly as I can in today’s video for you.

Comments are closed.