问题如下图：
选项：
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解释：
您好，请问这题中，d1算出来后，为什么代入的是N（-d1)，那个负号怎么理解呢？
感谢
NO.PZ2016082403000021 问题如下 A Europeput option hthe following characteristics: S0 =$50;X=$45;r=5%;T=1year;an=25%. Whiof the following is closest to the value of the put?Use the BSM formula to solve this question. N(-0.78644)=0.2158, N(-0.74644)=0.2266, N(-0.49644)=0.3085, N(-0.56644)=0.2855 A.$1.88. B.$3.28. C.$9.07. $10.39. S0 =$50; X=$45; r=5% T=1yeanσ=25% =ln(5045)+(0.05+0.06252)10.25(1)=0.186610.25=0.74644{\text{}_1=\frac{\ln(\frac{50}{45})+(0.05+\frac{0.0625}2)1}{0.25(1)}=\frac{0.18661}{0.25}=0.74644=0.25(1)ln(4550)+(0.05+20.0625)1=0.250.18661=0.74644=0.74644-0.25=0.49644from the cumulative normtableN(-)=0.2266N(-)=0.3085*p=45e−0.05(1)(0.3085)−50(0.2266)=1.88p=45e^{-0.05(1)}(0.3085)-50(0.2266)=1.88p=45e−0.05(1)(0.3085)−50(0.2266)=1.88(*note rounng) BSM模型中，put的lta是什么？
$3.28. $9.07. $10.39. A S0 =$50; X=$45; r=5% T=1yeanσ=25% =ln(5045)+(0.05+0.06252)10.25(1)=0.186610.25=0.74644{\text{}_1=\frac{\ln(\frac{50}{45})+(0.05+\frac{0.0625}2)1}{0.25(1)}=\frac{0.18661}{0.25}=0.74644=0.25(1)ln(4550)+(0.05+20.0625)1=0.250.18661=0.74644 =0.74644-0.25=0.49644 from the cumulative normtable N(-)=0.2266 N(-)=0.3085* p=45e−0.05(1)(0.3085)−50(0.2266)=1.88p=45e^{-0.05(1)}(0.3085)-50(0.2266)=1.88p=45e−0.05(1)(0.3085)−50(0.2266)=1.88 (*note rounng)请问可以给一下和公式吗 感觉和书上不太一样
请问，李老师讲课中说到put的bsm模型不用刻意去记，只要用c+k=p+s, 但是这里只给了n（-）和n（-），有没有办法处理，还是只能用put bsm的公式去做
为什么求put用负和负
