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台风来了 · 2024年04月08日

call和put的hedge ratio

NO.PZ2023041003000036

问题如下:

Suppose you observe a non- dividend- paying Australian equity trading for A$7.35. The call and put options have two years to mature, the periodically compounded risk- free interest rate is 4.35%, and the exercise price is A$8.0. Based on an analysis of this equity, the estimates for the up and down moves are u = 1.445 and d = 0.715, respectively.

Calculate the European- style call and put option hedge ratios at Time Step 0 and Time Step 1. Based on these hedge ratios, interpret the component terms of the binomial option valuation model.

选项:

解释:

The computation of the hedge ratios at Time Step 1 and Time Step 0 will require the option values at Time Step 1 and Time Step 2. The terminal values of the options are given in Solution 1.

S++ = u2S = 1.4452(7.35) = 15.347

S+– = udS = 1.445(0.715)7.35 = 7.594

S– – = d2S = 0.7152(7.35) = 3.758

S+ = uS = 1.445(7.35) = 10.621

S = dS = 0.715(7.35) = 5.255

Therefore, the hedge ratios at Time 1 are


In the last hedge ratio calculation, both put options are in the money (p–+ and p– –). In this case, the hedge ratio will be –1, subject to a rounding error. We now turn to interpreting the model’s component terms. Based on the no- arbitrage approach, we have for the call price, assuming an up move has occurred, at Time Step 1

C+ = hc+S+ + PV1,2 (- hc+S+- +C+-)-= 0.9476(10.621) + (1/1.0435)[–0.9476(7.594) + 0.0] = 3.1684

Thus, the call option can be interpreted as a leveraged position in the stock. Specifically, long 0.9476 shares for a cost of 10.0645 [= 0.9476(10.621)] partially financed with a 6.8961 {= (1/1.0435)[–0.9476(7.594) + 0.0]} loan. Note that the loan amount can be found simply as the cost of the position in shares less the option value [6.8961 = 0.9476(10.621) – 3.1684]. Similarly, we have

C-= hcS- + PV1,2 (- h cS-- +C--)-= 0.0(5.255) + (1/1.0435)[–0.0(3.758) + 0.0] = 0.0

Specifically, long 0.0 shares for a cost of 0.0 [= 0.0(5.255)] with no financing.

For put options, the interpretation is different. Specifically, we have

P+ =+ hP+S++ + PV1,2 (- hp+S++ +p++)-= (1/1.0435)[–(–0.05237)15.347 + 0.0] + (–0.05237)10.621 = 0.2140

Thus, the put option can be interpreted as lending that is partially financed with a short position in shares. Specifically, short 0.05237 shares for a cost of 0.55622 [= (–0.05237)10.621] with financing of 0.77022 {= (1/1.0435)[–(–0.05237)15.347 + 0.0]}. Note that the lending amount can be found simply as the proceeds from the short sale of shares plus the option value [0.77022 = (0.05237)10.621 + 0.2140]. Again, we have

P- = hPS- + PV1,2 (- hcS-+ +P-+)-= (1/1.0435)[–(–1.0)7.594 + 0.406] + (–1.0)5.255 = 2.4115

Here, we short 1.0 shares for a cost of 5.255 [= (–1.0)5.255] with financing of 7.6665 {= (1/1.0435)[–(–1.0)7.594 + 0.406]}. Again, the lending amount can be found simply as the proceeds from the short sale of shares plus the option value [7.6665 = (1.0)5.255 + 2.4115].

Finally, The interpretations remain the same at Time Step 0: c = hcS + PV0,1(–hcS + c) =


The interpretations remain the same at Time Step 0:

C=hcS+ PV0,1 (-hcS-+C-)-= 0.5905(7.35) + (1/1.0435)[–0.5905(5.255) + 0.0] = 1.37

Here, we are long 0.5905 shares for a cost of

4.3402 [=0.5905(7.35)] partially financed with a 2.97 {= (1/1.0435)[–0.5905(5.255) + 0.0] or = 0.5905(7.35) – 1.37} loan.

P = hPS+ PV0,1 (-hPS++P+)= (1/1.0435){–[–0.4095(10.621)] + 0.214} + (–0.4095)7.35 = 1.36

Here, we short 0.4095 shares for a cost of 3.01 [= (–0.4095)7.35] with financing of 4.37 (= (1/1.0435){–[–0.4095(10.621)] + 0.214} or = (0.4095)7.35 + 1.36).

老师,您好!


能不能解释一下为什么call和put的hedge ratio之差等于1 ? 即 hc - hp = 1 ?谢谢!

1 个答案

李坏_品职助教 · 2024年04月08日

嗨,爱思考的PZer你好:


hedge ratio其实就是期权的delta,看涨期权的delta是从0到1随着股价的上升而逐渐上升,看跌期权的delta是从-1到0随着股价的上升逐渐上升:

从BSM期权定价公式来看:


delta of call = d C / d S = N(d1),

而delta of put = d P / d S = -N(-d1),由标准正态分布的性质可得,-N(-d1) = -(1-N(d1)) = -1 + N(d1).

所以delta of call - delta of put = 1.

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NO.PZ2023041003000036 问题如下 Suppose you observe a non-vin paying Australiequity trang for A$7.35. The call anput optionshave two years to mature, the periocally compounrisk- free interest rateis 4.35%, anthe exercise priis A$8.0. Baseon analysis of this equity,the estimates for the up anwn moves are u = 1.445 an= 0.715,respectively. Calculatethe European- style call anput option hee ratios Time Step 0 anTimeStep 1. Baseon these hee ratios, interpret the component terms of thebinomioption valuation mol. The computation ofthe hee ratios Time Step 1 anTime Step 0 will require the option valuesTime Step 1 anTime Step 2. The terminvalues of the options are given inSolution 1. S++ = u2S= 1.4452(7.35) = 15.347 S+– =u = 1.445(0.715)7.35 = 7.594 S– – =S = 0.7152(7.35) = 3.758 S+ = uS= 1.445(7.35) = 10.621S– = = 0.715(7.35) = 5.255 Therefore, thehee ratios Time 1 are In the last heeratio calculation, both put options are in the money (p–+ anp– –). In thiscase, the hee ratio will –1, subjeto a rounng error. We now turn tointerpreting the mol’s component terms. Baseon the no- arbitrage approach,we have for the call price, assuming up move hoccurre Time Step 1• = hc+S+ + PV1,2 (- hc+S+- +C+-)-=0.9476(10.621) + (1/1.0435)[–0.9476(7.594) + 0.0] = 3.1684Thus, the calloption cinterpretea leverageposition in the stock. Specifically,long 0.9476 shares for a cost of 10.0645 [= 0.9476(10.621)] partially financeith a 6.8961 {= (1/1.0435)[–0.9476(7.594) + 0.0]} loan. Note ththe loanamount cfounsimply the cost of the position in shares less theoption value [6.8961 = 0.9476(10.621) – 3.1684]. Similarly, we have• C-= hc–S- + PV1,2 (- h c–S-- +C--)-= 0.0(5.255) + (1/1.0435)[–0.0(3.758) + 0.0] =0.0 Specifically, long0.0 shares for a cost of 0.0 [= 0.0(5.255)] with no financing. For put options,the interpretation is fferent. Specifically, we have P+ =+ hP+S+++ PV1,2 (- hp+S+++p++)-= (1/1.0435)[–(–0.05237)15.347 + 0.0] + (–0.05237)10.621 =0.2140 Thus, the putoption cinterpretelenng this partially financewith a shortposition in shares. Specifically, short 0.05237 shares for a cost of 0.55622 [=(–0.05237)10.621] with financing of 0.77022 {=(1/1.0435)[–(–0.05237)15.347 + 0.0]}. Note ththe lenng amount cbefounsimply the procee from the short sale of shares plus the optionvalue [0.77022 = (0.05237)10.621 + 0.2140]. Again, we have P- = hP–S-+ PV1,2 (- hc–S-++P-+)-= (1/1.0435)[–(–1.0)7.594 + 0.406] + (–1.0)5.255 = 2.4115 Here, we short 1.0shares for a cost of 5.255 [= (–1.0)5.255] with financing of 7.6665 {=(1/1.0435)[–(–1.0)7.594 + 0.406]}. Again, the lenng amount cfounimply the procee from the short sale of shares plus the option value[7.6665 = (1.0)5.255 + 2.4115]. Finally, Theinterpretations remain the same Time Step 0: c = h+ PV0,1(–hcS–+ c–) = Theinterpretations remain the same Time Step 0: C=hcS+PV0,1 (-hcS-+C-)-=0.5905(7.35) + (1/1.0435)[–0.5905(5.255) + 0.0] = 1.37 Here, we are long0.5905 shares for a cost of 4.3402 [=0.5905(7.35)] partially financewitha 2.97 {= (1/1.0435)[–0.5905(5.255) + 0.0] or = 0.5905(7.35) – 1.37} loan. P = hPS+PV0,1 (-hPS++P+)=(1/1.0435){–[–0.4095(10.621)] + 0.214} + (–0.4095)7.35 = 1.36Here, we short0.4095 shares for a cost of 3.01 [= (–0.4095)7.35] with financing of 4.37 (=(1/1.0435){–[–0.4095(10.621)] + 0.214} or = (0.4095)7.35 + 1.36). 这里求p-用的是7.594和0.406(有答案版本文件的37页最上面的公式),这两个数都是p+那个树杈上的,但是根据公式应该是用p-,3.757504和4,242496,为什么要用7.594和0.406?

2024-05-07 16:49 1 · 回答