Direct answer
A worked example of TSI (typically “True Strength Index”) shows how the indicator value is calculated step by step from a time series of prices. Below is a fully numerical example with explicit assumptions, so you can reproduce the same intermediate values and end result.
Mechanism or definition
TSI is built from (1) the price change (a momentum-like input) and (2) smoothing operations that reduce noise. Because “TSI” can be implemented in different ways across tools, a worked example must state the exact assumptions: definition, parameter values, and what data is used (such as Close-to-Close changes).
Assumptions used in this example (state them before calculating):
- We use a Close series and define the raw momentum as m[t] = Close[t] − Close[t−1].
- We use smoothing lengths r = 3 and s = 5 (small purely for demonstration; real settings often differ).
- We use simple moving averages (SMA) for smoothing, applied twice: first to the numerator and denominator components, then again to the results.
- Denominator uses absolute momentum: a[t] = |m[t]|.
- The TSI formula (with two-stage SMA) is:
- TSI[t] = 100 × ( SMA_s( SMA_r(m)[t] ) / SMA_s( SMA_r(a)[t] ) ).
Term meanings (plain language):
- Momentum m[t]: how much price changed since the last step.
- Two-stage smoothing: averaging the momentum, then averaging that average again to reduce short-term fluctuations.
- Scaling by 100: converts the ratio into an easier-to-read numeric range.
Evidence or example (complete numerical walk-through)
Given Close prices (10 points): Close[0..9] = [100, 101, 99, 102, 103, 101, 104, 105, 103, 106]
Step 1: Compute momentum and absolute momentum
Compute m[t] for t=1..9 and a[t]=|m[t]|:
- t1: m=101−100=1, a=1
- t2: m=99−101=−2, a=2
- t3: m=102−99=3, a=3
- t4: m=103−102=1, a=1
- t5: m=101−103=−2, a=2
- t6: m=104−101=3, a=3
- t7: m=105−104=1, a=1
- t8: m=103−105=−2, a=2
- t9: m=106−103=3, a=3
Step 2: First smoothing with SMA length r=3
For each t ≥ 3, compute:
- SMA_r(m)[t] = average of m over {t−2, t−1, t}
- SMA_r(a)[t] = average of a over {t−2, t−1, t}
Examples (we will need values around the point where we can also apply the second smoothing):
- For t=3: SMA3(m)[3]=(m1+m2+m3)/3=(1+(−2)+3)/3=2/3=0.6667 SMA3(a)[3]=(a1+a2+a3)/3=(1+2+3)/3=6/3=2.0000
- For t=4: SMA3(m)[4]=(m2+m3+m4)/3=(−2+3+1)/3=2/3=0.6667 SMA3(a)[4]=(a2+a3+a4)/3=(2+3+1)/3=6/3=2.0000
- For t=5: SMA3(m)[5]=(m3+m4+m5)/3=(3+1+(−2))/3=2/3=0.6667 SMA3(a)[5]=(a3+a4+a5)/3=(3+1+2)/3=6/3=2.0000
- For t=6: SMA3(m)[6]=(m4+m5+m6)/3=(1+(−2)+3)/3=2/3=0.6667 SMA3(a)[6]=(a4+a5+a6)/3=(1+2+3)/3=6/3=2.0000
- For t=7: SMA3(m)[7]=(m5+m6+m7)/3=(−2+3+1)/3=2/3=0.6667 SMA3(a)[7]=(a5+a6+a7)/3=(2+3+1)/3=6/3=2.0000
(Notice how this toy series makes SMA3(m) constant; that is not typical of real markets, but it keeps the demonstration transparent.)
Step 3: Second smoothing with SMA length s=5
TSI becomes available once we can compute SMA_s of the first-smoothed values. For s=5, we need five consecutive SMA_r(m) points.
To compute TSI[t] we use: TSI[t] = 100 × ( SMA5( SMA3(m) )[t] / SMA5( SMA3(a) )[t] )
Take t=7 (this is the first time we can form a 5-point SMA from SMA3 values starting at t=3 through t=7):
- SMA5( SMA3(m) )[7] = average of SMA3(m)[3..7] = average of five identical values 0.6667 = 0.6667
- SMA5( SMA3(a) )[7] = average of SMA3(a)[3..7] = average of five identical values 2.0000 = 2.0000