04 July Chart Decoder Series: Probabilistic Oscillator – Trader’s Radar for Reversal

Welcome to the Chart Decoder series, a guide to learning the essential tools to accurately read Bitfinex charts.

So far, I’ve learned how to spot trends (SMA/EMA), catch momentum shifts (MACD), and how to recognize excess/sold out zones (RSIs) and volatility extremes (bollinger bands).

So let’s talk about stochastic oscillators. This is an indicator of timing moments “too fast”;

What is a probabilistic oscillator?

Stochastic oscillators measure momentum by comparing the current closing price to the high and low ranges over a set period (usually 14 periods).

It consists of two rows.

• %k (high speed line): The main line responds immediately
• %D (slow line): Moving average %k that smooths noise

Typically, you get a reading between 0 and 100.

• Over 80 (%k): Too much bought. The market may need a breather
• Less than 20 (%k): Overselling. There may be a bounce
• Between 20 and 80:neutral. Prices have moved within recent ranges

Signal Crossover

• %k exceeds %d below 20: Bullish inverted signal
• %k is less than %d above 80: Bearish inverted signal
• If both lines are above 80 or below 20: It’s a strong trend, but it’s also a warning that you won’t get over the welcome. These zones often precede the shift.

Bitfinex BTC/USD example

• price:$102,150.
• Blue (k%):96.49
• Orange (D%):96.47
• There are both rows Excessive zone (Over 80). This often indicates the potential fatigue of current upward movements.

Pumps these days may be overdoing that. When a probabilistic measure makes this higher, it often shows it Buyers may be short of gasand it could be that the pullback is round the corner.

But here’s a twist. In particular, in crypto, excessive acquisitions do not mean “time to sell.” With strong trends, probabilistic momentum indicators may remain high for some time. So, you need more clues before you make a call.

What other metrics are useful?

Stochastic oscillators offer easy reading of extreme extremes, but work even better when paired with other tools.

Moving average of index (EMAS) It helps to zoom out, like 50 or 200. If the BTC is bouncing off, but still below 200 EMA, it could be a short-lived rescue rally, not a true reversal of the trend.

Macd It functions like a momentum compass. Although they say the probability theory is over-bought, if the MACD is gaining strength in the upward direction, that might mean there is room for more driving.

rsiOn the other hand, there are close probabilities. It also flags over and oversell zones, but has a smoother and slower approach. It’s perfect for checking if momentum is built or fading. When used together, RSI and probability theory give you both direction and timing. One shows where the market is tilting. The other one will tell you when you’re coming back.

RSI vs. Probability – What’s the difference?

Both indicate whether the coin is “over-acquired” or “over-selled”, but they work differently.

Bonus Read: When adding RSI to a mix

Layered to double check stochastic signals RSI (14) On the same BTC/USD chart. Here’s what we saw:

• Probabilistic It was over-buyed at 96+.
• rsi I sit comfortably 56.98to the right of the neutral zone.

This difference is interesting. Probability theory says “peak momentum,” but suggests that RSI is there There is still room for moving higher Before the trend is gone.

That discrepancy can often be shown Short-term indecisiveness Or even Opportunity to continueespecially when the volume starts.

So what’s moving? If the RSI breaks past 60 or 70 at a price-holding company, you may see this rally pushed further. However, if the RSI stalls and collapses probabilistically, it could be your early warning light.

Pro-use methods like probability theory:

• Checking is important: Don’t rely solely on probabilistic simultaneousness. Wait for the crossover and check the RSI or MACD.
• Don’t panic at every signal: It’s too much bought and can continue to be bought in a fierce bull market.
• Look for divergence: If the price is new and high but not probabilistic, momentum can fade and a reversal could come.
• Multiple time frames: A 15-minute chart signal makes little sense if every day is still trending.

Try it with Bitfinex:

1. Log in to Bitfinex
2. Please select a chart
3. Add a probabilistic oscillator
4. Beware of extreme crossovers

Next chart decoder series: VWAP and how to determine whether a price is above or below “fair value.”

Bitfinex. Original Bitcoin exchange.

]]> https://earlybirdsinvest.com/chart-decoder-series-probabilistic-oscillator-traders-radar-for-reversal/feed/ 0 46067 https://earlybirdsinvest.com/how-do-i-calculate-the-number-of-expected-sats-and-arrive-at-a-probabilistic-payment-flow/ https://earlybirdsinvest.com/how-do-i-calculate-the-number-of-expected-sats-and-arrive-at-a-probabilistic-payment-flow/#respond Sun, 06 Jul 2025 05:37:37 +0000 https://earlybirdsinvest.com/how-do-i-calculate-the-number-of-expected-sats-and-arrive-at-a-probabilistic-payment-flow/ Take a look at the following network example:

Assume S I want to send it 3 Sit down R. You can assume that further S Each local channel has sufficient liquidity 3 soil. It also assumes channel fluidity (A,R), (B,R) and (C,R) It is distributed evenly.

One optimally reliable payment flow in this diagram looks like this:

1 sat: S --> A --> R   probability: 2/3
2 sats: S --> B --> R  probability: 3/5

This flow has a total probability 2/3*3/5 = 2/5 = 0.4 = 40%

question:

How to calculate the expected value of the arrival of Satoshu R if S send 3?

Option a

(I already know I’m wrong, but I think some people have similar initial thoughts, so I’ll write it down)

At first I thought this was just right 3 sats * 2/5 = 6/5 sats = 1.2 sats This is obtained by multiplying the amount of transmission with the probability of flow. This seems strange when sending two SATs S-->B-->R There is a chance of 3/5 And with the above reasoning 2 sats * 3/5 = 6/5 sats = 1.2 sats. The expected value of 1 is sitting along S-->A-->B The path is bigger 0 This is inconsistent with the expected additive.

Option b

Starting with the above reasoning, we add the expected value to the broken path.

E(3 sats) = 1 sat * 2/3 + 2 sat * 3/5 = 10/15 sats + 18/15 sats = 28/15 sats

Option c

Of course, two Satoshi Passes S-->B-->R You don’t need to send it as a single onion as an onion, but you can send it as two onions each with one soil.

The first is the probability 4/5 The second is a conditional probability 3/4 This is widely explained in this issue. You should be able to add these expectations using the logic in option B. Therefore S--> B --> R It is calculated as follows:

E(2 sats) = 1 sat * 4/5 + 1 sat * 3/4 = 31/20 sats 

Add one Saturday onion S-->A-->R That was 2/3 soil

We expect to have

E(3 sats) = 31/20 sats + 2/3 sats = 93/60 sats + 40/60 sats = 132/60 sats = 33/15 sats

This is 5/15 sats = 1/3 sats More than the answer to option b

Option d

Worse, I’m confused if there’s a chance that the expectation of analyzing two SAT onions in option C into two SAT onions could be linearly added, as the second onion is conditional to have two SAT onions in the channel. If the first onion fails, the second onion certainly fails. Therefore, in order to send such two Saturday onions, you need to calculate the expected value.

E(2 sats) = 1 sat * 4/5 + 1 sat * 3/5 = 7/5 sats

This gives you the following total expected value:

E(3 sats) = 2/3 sats + 7/5 sats = 10/30 sats + 21/15 sats = 31/15 sats

thought

There are results just for comparison

• Option a: 18/15
• Option b: 28/15
• Option c: 33/15
• Option d: 31/15

Option B certainly seems correct, but it makes sense to further analyze the two SATS onions. In the simulation, option D appears to be correct. This is a bit surprising to me. Using the formalism of probability theory, the difference between the 2 SAT paths is as follows:

• Option c: E(2 sats) = 1 sat * P(X>=1) + 1 sat * P(X>=2 | X >= 1)
• Option d: E(2 sats) = 1 sat * P(X>=1) + 1 sat * P(X>=2)

As mentioned earlier, the simulated setting indicates that option D is correct, but that is very surprising to me as I expect the second term to be a conditional probability.