3. Parameterisation of the standardised interest rate shock scenarios

Given for each currency c the specified size of the parallel, short and long instantaneous shocks to the ‘risk-free’ interest rate, the following parameterisations of the six interest rate shock scenarios should be applied:

• (i) Parallel shock for currency c : A constant parallel shock up or down across all time buckets:
• (ii) Short rate shock for currency ∆𝑅𝑅 c : 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 Shock , up 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 or )= down ± 𝑅𝑅 � 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 that is greatest , 𝑐𝑐 at the shortest tenor

midpoint. That shock, through the shaping scalar −𝑡𝑡 k , where , diminishes toward zero at the tenor of the longest point on the term structure 𝑥𝑥²⁰ . Where t k is the midpoint (in time) of the k th bucket and t K is the midpoint 𝑆𝑆 𝑠𝑠ℎ𝑜𝑜𝑜𝑜𝑡𝑡 ( 𝑡𝑡 𝑘𝑘 (in ) = time) 𝑒𝑒 of the last 𝑥𝑥 bucket = 4

):

𝐾𝐾 k

• (iii) Long rate shock ∆𝑅𝑅 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 for , 𝑐𝑐 currency ( 𝑡𝑡 𝑘𝑘 )= ± c : 𝑅𝑅� This 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 shock , 𝑐𝑐 ∙ 𝑆𝑆 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 is only ( 𝑡𝑡 𝑘𝑘 applied )= ± 𝑅𝑅� to 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 rotational , 𝑐𝑐 ∙𝑒𝑒 𝑥𝑥 shocks. The shock is greatest at the longest tenor midpoint and is related to the short scaling factor as :

𝑆𝑆 𝑙𝑙𝑜𝑜𝑙𝑙𝑙𝑙 ( 𝑡𝑡 𝑘𝑘 ) = 1 − 𝑆𝑆 𝑠𝑠ℎ𝑜𝑜𝑜𝑜𝑡𝑡 ( 𝑡𝑡 𝑘𝑘 ) k

• (iv) Rotation shocks ∆𝑅𝑅 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 for , 𝑐𝑐 ( 𝑡𝑡 currency 𝑘𝑘 )= ± 𝑅𝑅� c 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 : Involving , 𝑐𝑐 ∙ 𝑆𝑆 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 rotations ( 𝑡𝑡 𝑘𝑘 )= ± to 𝑅𝑅� the 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 term , 𝑐𝑐 ∙� structure 1 −𝑒𝑒 𝑥𝑥 � (i.e. for steepeners

and flatteners ) of the interest rates, whereby both the long and short rates are shocked and the shift in interest rates at each tenor midpoint is obtained by applying the following formulas to those shocks:

∆𝑅𝑅 𝑠𝑠𝑜𝑜𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑙𝑙𝑝𝑝𝑝𝑝 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 ) = − 0.65 ∙ ∣ ∆𝑅𝑅 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 ) ∣ +0.9 ∙ ∣ ∆𝑅𝑅 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 ) ∣ Examples: ∆𝑅𝑅 𝑓𝑓𝑝𝑝𝑝𝑝𝑜𝑜𝑜𝑜𝑝𝑝𝑙𝑙𝑝𝑝𝑝𝑝 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 )= +0.8 ∙ ∣ ∆𝑅𝑅 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 ) ∣ − 0.6 ∙ ∣ ∆𝑅𝑅 𝑝𝑝𝑜𝑜𝑙𝑙𝑙𝑙 , 𝑐𝑐 ( 𝑡𝑡 𝑘𝑘 ) ∣ .

Short rate shock : Assume the bank uses time bands and t K = 25 years (the midpoint (in time) of the longest tenor bucket ), where is the midpoint (in time) for bucket k. For with t k 𝐾𝐾 = 19 = 3.5 years, the scalar adjustment 𝐾𝐾 for the 𝑡𝑡 𝑘𝑘 short shock would be: −3 .⁵ = 𝑘𝑘 0.417. = 10 Banks would multiply this by the value of the short rate shock to obtain the amount⁴ to be added or subtracted from the yield curve at that tenor point. If the short 𝑆𝑆 𝑠𝑠ℎ𝑜𝑜𝑝𝑝𝑜𝑜 rate ( 𝑡𝑡 shock 𝑘𝑘 )= 𝑒𝑒 was +250 bps, the

increase in the yield curve at t k = 3.5 years would be 104.2 bps.

Steepener: Assume the same point on the yield curve as above, t k = 3.5 years. If the absolute value of the short rate shock was 250 bps and the absolute value of the long rate shock was 100 bps (as for the euro), the change in the yield curve at t k = 3.5 years would be the sum of the effect of the short rate shock plus the effect of the long rate shock in basis points: . − 0.65 ∙ 250 bps ∙ 0.417+ 0.9 ∙ 100 bps ∙ (1 − 0.417)= − 15.3 bps Flattener: The corresponding change in the yield curve for the shocks in the example above at t k = 3.5 years would be: .

+ 0.8 ∙ 250 bps ∙ 0.417 − 0.6 ∙ 100 bps ∙ (1 − 0.417)= 48.4 bps

²⁰ The value of in the denominator of the function −𝑡𝑡 k controls the rate of decay of the shock. 𝑥𝑥 𝑥𝑥 𝑒𝑒 51

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