Strike Price Convexity: Practice Problems and Solutions

This is Section 12 of the Study Guide. See Section 1 here. See Section 2 here. See Section 3 here. See Section 4 here. See Section 5 here. See Section 6 here. See Section 7 here. See Section 8 here. See Section 9 here. See Section 10 here. See Section 11 here.

For strike prices K₁< K₂< K₃, the following inequalities express the convexity of the option price with regard to the strike price:

[C(K₁) - C(K₂)]/[K₂ - K₁] ≥ [C(K₂) - C(K₃)]/[K₃ - K₂]

[P(K₂) - P(K₁)]/[K₂ - K₁] ≤ [P(K₃) - P(K₂)]/[K₃ - K₂]

Where C(K_x) and P(K_x) are prices of American or European options with strike price K_x.

It is also possible to define λ = [K₃ - K₂]/[K₃ - K₁] and rewrite the inequality for convexity with calls as

C(K₂) ≤ λC(K₁) + (1- λ)C(K₃)

A similar relationship holds for puts:

P(K₂) ≤ λP(K₁) + (1- λ)P(K₃)

Original Practice Problems and Solutions from the Actuary's Free Study Guide:

Problem SPC1. Vacuous Co. stock options trade for three strike prices: $46, $32, and $90. Determine λ in the inequality for strike price convexity.

Solution SPC1. λ = [K₃ - K₂]/[K₃ - K₁] Here, K₁ = 32, K₂ = 46, and K₃ = 90.

Thus, λ = [90- 46]/[90- 32] = λ = 0.7586296897

Problem SPC2. Call options on Voracious, Inc., stock trade for three strike prices: $43, $102, and $231. The price of the $43-strike option is $56. The price of the $231-strike option is $23. What is the maximum possible price of the $102-strike option? All options have the same time to expiration.

Solution SPC2. First, we find λ = [K₃ - K₂]/[K₃ - K₁]. Here, K₁ = 43, K₂ = 102, and K₃ = 231. Thus, λ = [231- 102]/[231- 43] = 0.6861702128.

We are given that C(K₁) = 56 and C(K₃) = 23. Thus, by the inequality

C(K₂) ≤ λC(K₁) + (1- λ)C(K₃), C(K₂) ≤ 0.6861702128*56 + (1- 0.6861702128)*23 = 45.64361702. So the maximum price of the $102-strike option is $45.64361702.

Problem SPC3. Call options on Specious LLC stock trade for three strike prices, $32, $34, and $23. The $32-strike call currently costs $10, while the $34-strike call costs $7. The $23-strike call costs at least $X. Find X.

Solution SPC3. First, we find λ = [K₃ - K₂]/[K₃ - K₁]. Here, K₁ = 23, K₂ = 32, and K₃ = 34. Thus, λ = [34- 32]/[34- 23] = 0.1818181818. We rearrange the inequality

C(K₂) ≤ λC(K₁) + (1- λ)C(K₃) to look as follows:

C(K₂) - (1- λ)C(K₃) ≤ λC(K₁)

[C(K₂) - (1- λ)C(K₃)]/λ ≤ C(K₁)

Thus, C(K₁) ≥ [10 - (1- 0.1818181818)7]/0.1818181818 = 23.5. Thus, the $23-strike call costs at least $23.5.

Problem SPC4. Put options on Meritorious Co. trade for three strike prices: $102, $105, and X, which is the highest. λ is equal to 0.5. The $102-strike put is worth $20, the $105 strike put is worth $22, and the $X-strike put is worth $24. Find the value of X.

Solution SPC4. This problem contains a lot of excess information.

λ = [K₃ - K₂]/[K₃ - K₁], so all we need to know is that λ = 0.5, K₁ = 102, and K₂ = 105.

Thus, 0.5 = (X - 105)/(X-102). Hence, 0.5X - 51 = X - 105 and 0.5X = 54.

Therefore, X = $108.

Problem SPC5. Put options on Meritorious Co. trade for three strike prices: $102, $105, and $112. The $102-strike put is worth $20, the $105 strike put is worth $22, and the $112-strike put is worth at least $F. Find the value of F.

Solution SPC5. Here, K₁ = 102, K₂ = 105, and K₃ = 112.

λ = [K₃ - K₂]/[K₃ - K₁] = (112-105)/(112-102) = 0.7

We rearrange the inequality P(K₂) ≤ λP(K₁) + (1- λ)P(K₃) to look as follows:

P(K₂) - λP(K₁) ≤ (1- λ)P(K₃)

[P(K₂) - λP(K₁)]/(1- λ) ≤ P(K₃)

[22 - 0.7*20]/0.3 = 26.66666667 ≤ P(K₃). So F = 26.66666667.

See other sections of The Actuary's Free Study Guide for Exam 3F / Exam MFE.