1 00:00:00,990 --> 00:00:04,730 Okay, and now I think we have sort of the 2 00:00:04,730 --> 00:00:09,730 the, tools we needed, to get to the the important point. 3 00:00:09,730 --> 00:00:12,370 Which was differentiating improper integrals. 4 00:00:21,470 --> 00:00:23,830 So we're going to start out with a function F. 5 00:00:23,830 --> 00:00:29,370 That's continuous, and we're going to assume that the integral of f over 6 00:00:29,370 --> 00:00:34,920 the entire real number line, so integral from minus infinity to infinity, exists. 7 00:00:34,920 --> 00:00:37,490 So, in particular, that means that it's finite. 8 00:00:43,790 --> 00:00:48,660 And now I want to consider a function b of t. 9 00:00:48,660 --> 00:00:50,940 Its a differentiable function and I am going to 10 00:00:50,940 --> 00:00:54,220 define a new function g of t which is 11 00:00:54,220 --> 00:01:00,430 going to be the integral of my function f from minus infinity up to b of t, 12 00:01:05,290 --> 00:01:07,498 and I want to be able to compute g prime of t. 13 00:01:09,168 --> 00:01:14,150 And so so, sort of like taking the derivative of a definite integral. 14 00:01:14,150 --> 00:01:17,700 It's actually, doesn't end up being too difficult. 15 00:01:17,700 --> 00:01:21,930 So all we're going to do, because we assumed that this integral 16 00:01:21,930 --> 00:01:27,000 exists, that means that I can pick a point somewhere in between. 17 00:01:28,890 --> 00:01:30,450 And I can write g 18 00:01:30,450 --> 00:01:33,900 of t as an integral from minus infinity up to that 19 00:01:33,900 --> 00:01:39,160 point plus the integral from that point up to b of t. 20 00:01:40,410 --> 00:01:44,840 And then because I'm free to choose that point I might as 21 00:01:44,840 --> 00:01:47,403 well choose something that's easy to work with so I chose 0. 22 00:01:51,440 --> 00:01:56,765 And so now I just want to take the derivative of G of T, but 23 00:01:56,765 --> 00:02:02,140 since the first term, 24 00:02:02,140 --> 00:02:06,780 so if I look at this closely now, this is a, it's an indefinite integral. 25 00:02:07,880 --> 00:02:11,280 So the integral from minus infinity up to zero. 26 00:02:11,280 --> 00:02:16,490 But there's no t's in this first term. So as far as I'm concerned, 27 00:02:16,490 --> 00:02:21,180 when I'm taking a derivative with respect to the variable t, this is a constant. 28 00:02:24,650 --> 00:02:26,340 And so, when I take the derivative of that 29 00:02:26,340 --> 00:02:28,790 with respect to t, when I take its derivative. 30 00:02:28,790 --> 00:02:29,950 I'm just going to get zero. 31 00:02:30,990 --> 00:02:35,740 So, to compute the derivative with respect to t of my function 32 00:02:35,740 --> 00:02:40,300 g of t, I can just ignore this first term and all I 33 00:02:40,300 --> 00:02:43,720 have to do is compute the derivative with respect to t of the 34 00:02:43,720 --> 00:02:47,879 integral from zero to b of t of my function f of x. 35 00:02:50,830 --> 00:02:52,960 But notice what happened now. 36 00:02:52,960 --> 00:02:55,060 So the, the problem that we were trying to deal 37 00:02:55,060 --> 00:02:57,810 with is that we had an, we had an improper integral. 38 00:02:59,140 --> 00:03:05,470 But because of this the limits here are both finite values. 39 00:03:05,470 --> 00:03:09,460 So, this is integral that we already know how to take the derivative of. 40 00:03:09,460 --> 00:03:12,240 This is just a, take the, we use exactly the same rule 41 00:03:12,240 --> 00:03:16,292 I talked about in I guess it would be number seven today. 42 00:03:16,292 --> 00:03:19,260 How to take the derivative of a definite integral. 43 00:03:19,260 --> 00:03:23,020 And, on top of that, we're lucky that the bottom is 44 00:03:23,020 --> 00:03:25,700 zero so we only have to use half of the formula. 45 00:03:28,730 --> 00:03:34,960 So, g prime of t is just going to be the derivative with respect to t of the 46 00:03:34,960 --> 00:03:37,870 anti derivative of f evaluated at b of t 47 00:03:37,870 --> 00:03:41,010 minus the anti derivative of f evaluated at zero. 48 00:03:42,090 --> 00:03:46,850 And this is just the result from differentiating. 49 00:03:46,850 --> 00:03:48,380 a definite integral. 50 00:03:48,380 --> 00:03:54,040 So I know that that's f of b of t times the derivative of b of t. 51 00:03:54,040 --> 00:03:59,710 So f of b of t times b prime of t. 52 00:03:59,710 --> 00:04:03,360 And then to do the same thing for the lower limit. 53 00:04:04,720 --> 00:04:06,780 Or we can do the same thing for the lower limit. 54 00:04:06,780 --> 00:04:12,410 So I'm just going to instead of g of t, I'm going to have h of t, just exactly the 55 00:04:12,410 --> 00:04:14,450 same thing, except that now it's my lower limits 56 00:04:14,450 --> 00:04:16,510 that's the function of t rather than the upper limit, 57 00:04:16,510 --> 00:04:17,010 [NOISE]. 58 00:04:24,300 --> 00:04:29,631 And so I'm still going to consider integrating this function from a of t to 59 00:04:29,631 --> 00:04:35,010 zero, and then adding on the integral from zero to infinity. 60 00:04:37,890 --> 00:04:40,450 And then when I take the derivative of that, this 61 00:04:40,450 --> 00:04:43,230 second term now is going to be the constant, so that's 62 00:04:43,230 --> 00:04:45,480 just going to go away so the, the answer here 63 00:04:45,480 --> 00:04:47,680 is just going to be the derivative of this first term. 64 00:04:51,000 --> 00:04:55,410 And so the only difference we're going to see, is just because 65 00:04:55,410 --> 00:04:58,630 when I do this part of the fundamental theorem of Calculus. 66 00:04:58,630 --> 00:05:02,420 Now my function a of t is on the negative side, so I get 67 00:05:02,420 --> 00:05:07,980 the same answer as I got for the, for the upper, limit being the function. 68 00:05:07,980 --> 00:05:09,635 Except now it's just going to be minus. 69 00:05:09,635 --> 00:05:16,080 F of a of t times a prime of t. So now we have a, a very simple 70 00:05:16,080 --> 00:05:19,220 rule for taking the derivative of an indefinite integral. 71 00:05:19,220 --> 00:05:22,250 And let's see if we can put that to work. 72 00:05:23,465 --> 00:05:26,565 So, 73 00:05:26,565 --> 00:05:26,590 [NOISE] 74 00:05:26,590 --> 00:05:30,780 I'm going to look at something called the delta of a European Call Option. 75 00:05:30,780 --> 00:05:33,280 That I'm going to price using the Black-Scholes formula. 76 00:05:35,120 --> 00:05:38,810 And so the Black-Scholes price, for a European call option, 77 00:05:38,810 --> 00:05:41,696 so this is the simplest possible one to look at. 78 00:05:41,696 --> 00:05:46,530 On a non-dividend paying asset, so this could just be 79 00:05:46,530 --> 00:05:49,842 a stock for instance, with a strike price of 1. 80 00:05:51,720 --> 00:05:56,160 Has a, a volatility of 20%, so this means the volatility of 81 00:05:56,160 --> 00:06:00,640 the returns on my asset, and I'm looking at three months to maturity. 82 00:06:00,640 --> 00:06:04,030 And so again, the reason I'm, I'm putting all of these real numbers in 83 00:06:04,030 --> 00:06:06,030 here is just because the Black-Scholes pricing 84 00:06:06,030 --> 00:06:09,540 formula, it's a function of several variables. 85 00:06:09,540 --> 00:06:12,200 I'm going to make it into a function of a single 86 00:06:12,200 --> 00:06:15,480 variable s which is the, the current price of the asset, 87 00:06:16,750 --> 00:06:21,540 just by plugging in actual numeric values to all of the rest of the inputs. 88 00:06:23,188 --> 00:06:25,120 And so this is the, the pricing formula. 89 00:06:26,320 --> 00:06:30,900 So I have s. Times this function phi, evaluated at d 90 00:06:30,900 --> 00:06:36,320 1, minus e to the minus 0.01, 91 00:06:36,320 --> 00:06:41,560 times this function phi again, evaluated at d 2. 92 00:06:41,560 --> 00:06:45,860 Where d 1 of S is just going to equal 10 93 00:06:45,860 --> 00:06:50,270 times the log of the, the current stock price, so 10 times the log of S. 94 00:06:50,270 --> 00:06:58,990 Plus 0.015. D2 of s is equal to d1 of s minus 0.1. 95 00:06:58,990 --> 00:07:02,100 And this function phi. 96 00:07:02,100 --> 00:07:06,820 So this is where the, all of the math we've been doing 97 00:07:06,820 --> 00:07:09,160 for the, for the last two lectures is going to come into play. 98 00:07:09,160 --> 00:07:09,760 This function phi. 99 00:07:10,880 --> 00:07:12,950 Is defined as the integral. 100 00:07:12,950 --> 00:07:15,650 So this is a, a capital phi, wherein the font 101 00:07:15,650 --> 00:07:19,120 I'm using, it's drawn just with a vertical middle line. 102 00:07:19,120 --> 00:07:25,080 The lowercase phi is its derivative, which has a, a slightly slanted line. 103 00:07:25,080 --> 00:07:29,990 So I'm going to integrate lowercase phi from minus infinity up to z. 104 00:07:32,220 --> 00:07:37,060 And lower case v this is probability density function for a 105 00:07:37,060 --> 00:07:42,020 standard normal distribution and this is the mathematical formula it has. 106 00:07:43,700 --> 00:07:46,640 So in order to calculate the price of my European call 107 00:07:46,640 --> 00:07:50,500 option, I am going to need to be able to do mathematics. 108 00:07:50,500 --> 00:07:54,240 With this function capital phi. 109 00:07:55,750 --> 00:07:57,450 But this is a function that's sort of famous 110 00:07:57,450 --> 00:07:59,700 for not having an antiderivative. 111 00:07:59,700 --> 00:08:04,800 So you can't actually write down a formula for the value of this integral. 112 00:08:04,800 --> 00:08:08,040 This is the simplest that I can write it down. 113 00:08:08,040 --> 00:08:10,340 So in order to actually price these things, you have 114 00:08:10,340 --> 00:08:13,380 to use a computer to approximate the value of this integral. 115 00:08:15,550 --> 00:08:18,930 But when we do mathematics with the Black-Scholes formula, 116 00:08:18,930 --> 00:08:21,110 we still want to be able to get analytical results. 117 00:08:21,110 --> 00:08:21,290 You know? 118 00:08:21,290 --> 00:08:24,500 If, if I try to do everything numerically. 119 00:08:24,500 --> 00:08:28,830 Then you run into problems of, you have to be very careful of, you know? 120 00:08:28,830 --> 00:08:31,100 As you continue to do calculations, are you 121 00:08:31,100 --> 00:08:32,910 accumulating any round off error in your compu 122 00:08:32,910 --> 00:08:34,150 [INAUDIBLE]. 123 00:08:34,150 --> 00:08:37,760 In your sums things like that, so what we would like 124 00:08:37,760 --> 00:08:41,428 to be able to do is actually compute derivatives of the 125 00:08:41,428 --> 00:08:46,620 Black-Scholes pricing formula and get the correct answer theoretically before we 126 00:08:46,620 --> 00:08:50,850 actually have to start plugging in numbers to the to the formula. 127 00:08:53,580 --> 00:08:58,960 And so, the delta of a European call option is just the derivative of the 128 00:08:58,960 --> 00:09:04,160 price, so the derivative of the pricing formula, so in this case c of s with 129 00:09:04,160 --> 00:09:08,980 respect to the asset price s. And it's denoted by 130 00:09:08,980 --> 00:09:13,890 this Equilateral triangle is capital Greek letter delta. 131 00:09:16,170 --> 00:09:18,750 And so this is the derivative that I'm going to try and compute. 132 00:09:21,800 --> 00:09:28,490 So, it's already kind of complicated and it's much, much 133 00:09:28,490 --> 00:09:31,340 more complicated if I don't plug numbers into it, because 134 00:09:31,340 --> 00:09:33,010 as soon as I plug the numbers into it, I 135 00:09:33,010 --> 00:09:36,046 can do some math and a lot of them combine together. 136 00:09:36,046 --> 00:09:39,780 but it's still complicated enough that I'm going to try 137 00:09:39,780 --> 00:09:42,060 and compute the derivative of this term by term. 138 00:09:42,060 --> 00:09:44,530 So first I'll look at this term and then we'll 139 00:09:44,530 --> 00:09:47,190 look at this term and we'll try and simplify our answers 140 00:09:47,190 --> 00:09:47,890 as much as we can. 141 00:09:50,590 --> 00:09:56,840 So for the derivative of the first term I have S times a function of S. 142 00:09:56,840 --> 00:09:58,530 So I'm going to need to use the product rule. 143 00:09:58,530 --> 00:10:01,270 So that's just going to tell me that the derivative of this 144 00:10:01,270 --> 00:10:06,020 function here is going to be the derivative of s times this. 145 00:10:06,020 --> 00:10:08,650 Plus the derivative of this times s. 146 00:10:12,190 --> 00:10:15,780 So here, I have the derivative of s is just, oops. 147 00:10:15,780 --> 00:10:17,860 Derivative of s is just 1. 148 00:10:17,860 --> 00:10:19,940 So I just have the second piece here. 149 00:10:21,610 --> 00:10:27,910 Then plus s times the derivative of, this guy. 150 00:10:27,910 --> 00:10:30,090 And I'm just going to replace that by its definition. 151 00:10:30,090 --> 00:10:33,080 So that's the, the definition of capital phi of d one of 152 00:10:33,080 --> 00:10:38,010 s is the integral from minus infinity up to d one of s. 153 00:10:38,010 --> 00:10:39,700 Of little phi of x. 154 00:10:45,300 --> 00:10:50,290 And so we have a formula now for evaluating this derivative. 155 00:10:51,470 --> 00:10:54,740 So, if I'm taking the derivative of an indefinite 156 00:10:54,740 --> 00:10:58,309 integral, I know that it's just going to be. 157 00:10:59,930 --> 00:11:04,610 The integrand evaluated at the upper limit times the derivative of 158 00:11:04,610 --> 00:11:07,890 the upper limit, with respect to the variable that I'm, I'm computing. 159 00:11:07,890 --> 00:11:09,330 So I need to use the chain rule here. 160 00:11:10,540 --> 00:11:12,710 And so I'm going to end up with s, that's this 161 00:11:12,710 --> 00:11:17,040 s here, times lower case phi of d 1 of s. 162 00:11:17,040 --> 00:11:20,360 So that's this phi evaluated at the upper end point of the integral 163 00:11:21,620 --> 00:11:25,380 and then by the chain rule this is phi of d of s. 164 00:11:25,380 --> 00:11:30,220 So I need to also multiply that by the derivative of d1 of s. 165 00:11:34,630 --> 00:11:38,649 So now remember that d1 of s is equal to 10 times the log of s. 166 00:11:40,240 --> 00:11:46,530 Plus 0.015, so the derivative with respect to s of d1 of s 167 00:11:46,530 --> 00:11:52,150 is just going to be 10 divided by s, so 10 times this here that's just a 168 00:11:52,150 --> 00:11:56,550 constant so the derivative of the constant is zero, the derivative of the log of 169 00:11:56,550 --> 00:11:59,680 s with respect to s is just 1 over s so I end up with 170 00:11:59,680 --> 00:12:00,549 10 divided by s. 171 00:12:04,600 --> 00:12:09,754 So that tells me that the derivative of the first term now, this capital phi 172 00:12:09,754 --> 00:12:15,370 of d1 of s plus s times lower-case phi of d1 of s, 173 00:12:15,370 --> 00:12:20,112 times 10 over s, and luckily one of the nice things about Black-Scholes is 174 00:12:20,112 --> 00:12:25,600 usually when you're doing it right stuff kind of beneficially cancels out for you. 175 00:12:25,600 --> 00:12:29,250 So here, the s and 1 over s are going to cancel each other out. 176 00:12:30,500 --> 00:12:34,420 So for the first term, the derivative is capital phi of d 177 00:12:34,420 --> 00:12:38,620 1 of S, plus 10 times little phi of d 1 of S. 178 00:12:41,460 --> 00:12:41,730 Okay. 179 00:12:41,730 --> 00:12:45,710 So now let's look at the derivative of the second term. 180 00:12:45,710 --> 00:12:49,460 So here it's a little bit easier, so e to the minus 0.01. 181 00:12:49,460 --> 00:12:52,630 Looks complicated but that's just a constant. 182 00:12:52,630 --> 00:12:54,050 [COUGH] 183 00:12:54,050 --> 00:12:57,080 So I'm going to take that outside of 184 00:12:57,080 --> 00:13:00,170 my derivative, so the derivative was linear remember. 185 00:13:00,170 --> 00:13:07,610 And I'm left with the derivative of capital phi of d2 of s. 186 00:13:07,610 --> 00:13:10,910 And I'm just going to, so in the square brackets is just the definition of that. 187 00:13:10,910 --> 00:13:16,580 So that's the integral from minus infinity to d2 of s, of little phi of x dx. 188 00:13:22,300 --> 00:13:28,310 And so I'm going to use my pricing rule sorry my im. 189 00:13:28,310 --> 00:13:29,650 Improper integral rule 190 00:13:31,660 --> 00:13:32,170 again. 191 00:13:32,170 --> 00:13:36,470 So that's just going to be phi evaluated at 192 00:13:36,470 --> 00:13:40,290 the upper endpoint, and then by the chain rule, 193 00:13:40,290 --> 00:13:42,300 I need to divide that, or, sorry, I need 194 00:13:42,300 --> 00:13:45,389 to multiply that by the derivative of d2 of S. 195 00:13:49,470 --> 00:13:53,930 And so remember, d2 of s was just equal to d1 of s minus 0.1. 196 00:13:53,930 --> 00:13:54,430 So 197 00:13:57,410 --> 00:14:00,490 this is just a constant, so the derivative of d2 with respect 198 00:14:00,490 --> 00:14:03,050 to s has to be the equal to the derivative of d1 199 00:14:03,050 --> 00:14:07,030 with respect to s, and we already computed that, and that was 200 00:14:07,030 --> 00:14:09,690 10 over s, so we can recycle that result a little bit. 201 00:14:12,320 --> 00:14:17,760 So that's this parked over here I can replace with 10 over S and 202 00:14:17,760 --> 00:14:22,920 that gives me e to the minus 0.01 times phi of 203 00:14:22,920 --> 00:14:27,950 d 2 of S times 10 over S. And 204 00:14:29,530 --> 00:14:33,613 so if I put all of that together. The derivative of the first term 205 00:14:33,613 --> 00:14:37,506 was capital phi of d 1 plus 206 00:14:37,506 --> 00:14:41,128 10 times lower case phi of d 1 of s. 207 00:14:43,230 --> 00:14:45,110 And the derivative of the second term so I was 208 00:14:45,110 --> 00:14:48,220 subtracting the second term so it's going to be subtracted here. 209 00:14:48,220 --> 00:14:49,325 I have10 over s. 210 00:14:49,325 --> 00:14:55,070 E to the minus 0.01 times phi of d 2 of S. 211 00:14:55,070 --> 00:14:57,180 So sorry, I should, this d 1 is a function of 212 00:14:57,180 --> 00:14:59,740 S, so it should have an S in the parentheses, as well. 213 00:14:59,740 --> 00:15:00,710 I forgot to put that there. 214 00:15:07,050 --> 00:15:07,050 [COUGH] 215 00:15:07,050 --> 00:15:09,240 But unfortunately, we're not finished yet. 216 00:15:11,770 --> 00:15:17,980 So remember that little phi of x is equal to 1 over the square root of 2 pi times 217 00:15:17,980 --> 00:15:24,940 e to the minus x squared over 2. So 218 00:15:24,940 --> 00:15:29,970 what I'd like to do to simplify this. Instead of having a d1 and a d2 in here. 219 00:15:29,970 --> 00:15:32,410 I want to see if I can rewrite, just remember d, 220 00:15:32,410 --> 00:15:36,120 d1 and d2 weren't that different, d2 was just d1 minus 0.1. 221 00:15:36,120 --> 00:15:38,100 So maybe 222 00:15:38,100 --> 00:15:42,620 I can make this simpler by writing writing this term 223 00:15:42,620 --> 00:15:45,079 in terms of d1 of S rather than d2 of S. 224 00:15:49,370 --> 00:15:53,883 So I'm just going to start plugging things into this definition of lower case phi. 225 00:15:56,780 --> 00:15:59,592 And so first of all, I'm just going to evaluate phi 226 00:15:59,592 --> 00:16:03,370 at d 2 so I have d 2 of s squared. 227 00:16:03,370 --> 00:16:05,300 Where they used to just be the argument here. 228 00:16:08,290 --> 00:16:11,010 But I want to have a d 1 there instead. 229 00:16:11,010 --> 00:16:15,950 So I'm going to replace my d 2 with d 1 of S, minus 0.1. 230 00:16:15,950 --> 00:16:21,730 So I have e to the minus phi 1 of S minus 0.1, that quantity squared. 231 00:16:24,370 --> 00:16:27,270 And now I'm going to try and to, to make a little bit of sense of this. 232 00:16:27,270 --> 00:16:32,500 It would be nice if I had this just in terms of e to the 233 00:16:32,500 --> 00:16:37,520 d1 of S squared because that, that would allow me to write it in terms 234 00:16:37,520 --> 00:16:42,569 of phi of d1 of S which would make it match what's on in this second term here. 235 00:16:44,060 --> 00:16:47,630 And so now what I want to do is sort of my completing the squares trick backwards, 236 00:16:50,040 --> 00:16:55,310 so let's just start by doing the math. So the first thing I'm going to do is 237 00:16:55,310 --> 00:17:01,010 expand the, the squared term here. So I had. 238 00:17:01,010 --> 00:17:07,330 D1 of S minus 0.1 squared. So that ends up being d1 squared of S 239 00:17:07,330 --> 00:17:14,144 minus 2 times 0.1 times d1 of S, so I have 0.2 d1 of S. 240 00:17:14,144 --> 00:17:14,646 Then, 241 00:17:14,646 --> 00:17:21,191 plus negative 0.1 squared, so that becomes positive 0.01. 242 00:17:27,070 --> 00:17:27,630 Let's see. 243 00:17:31,420 --> 00:17:33,780 Did anything change? No. 244 00:17:33,780 --> 00:17:37,340 I think I made a cutting and pasting error in this one, so it's hope, hopefully, that 245 00:17:37,340 --> 00:17:44,894 doesn't last too long. Okay, so essentially what was 246 00:17:44,894 --> 00:17:51,836 supposed to happen in this line, 247 00:17:51,836 --> 00:17:58,520 and I guess I somehow missed it. 248 00:17:58,520 --> 00:18:03,635 Here I have 0.01 divided by 2, so that's 249 00:18:03,635 --> 00:18:08,590 going to be 0.005 and remember when 250 00:18:08,590 --> 00:18:14,190 I'm adding things together in an exponent, I can also think of that as a product 251 00:18:14,190 --> 00:18:18,980 of e to each one of those terms. So what I meant to 252 00:18:18,980 --> 00:18:23,630 do in this line here, I wanted to have e to 253 00:18:23,630 --> 00:18:27,970 the minus d1 squared over 2. That's what's going to end up here. 254 00:18:29,120 --> 00:18:34,100 I'm going to have minus times minus, so. 255 00:18:34,100 --> 00:18:38,180 E to the minus, or e to the positive 0.2 d one but 256 00:18:38,180 --> 00:18:41,790 I'm dividing that by two so I get e to the 0.1 d one. 257 00:18:41,790 --> 00:18:42,290 And 258 00:18:44,380 --> 00:18:51,240 then, 0.01 divided by two was 0.05. And that was just going to end up also 259 00:18:51,240 --> 00:18:55,398 over here and remember there's the minus sign multiplying it here so I have minus. 260 00:18:55,398 --> 00:19:01,170 0.01 minus 0.005. And that's what gives me this term here. 261 00:19:03,600 --> 00:19:05,250 So hopefully, that would have been a little bit 262 00:19:05,250 --> 00:19:08,790 more clear if I had, not messed up this line. 263 00:19:08,790 --> 00:19:09,290 But, 264 00:19:12,690 --> 00:19:15,150 okay. And now I've, I've written this bit. 265 00:19:15,150 --> 00:19:17,050 The one over 2 pi here. 266 00:19:17,050 --> 00:19:22,020 E to the minus d 1 of s squared over 2, 267 00:19:22,020 --> 00:19:24,930 well that's exactly what phi of d 1 of s would be. 268 00:19:24,930 --> 00:19:29,210 So hopefully that's starting to look a little bit simpler now. 269 00:19:30,870 --> 00:19:35,509 And now remember I had d 1 was just 10 times. 270 00:19:37,900 --> 00:19:42,520 The log of, 10 times the log of S, hopefully that'll show 271 00:19:42,520 --> 00:19:46,988 up here. 10 times the log of S plus 272 00:19:46,988 --> 00:19:52,110 0.15. And so 273 00:19:52,110 --> 00:19:57,820 here I have e to the log of S. That's just going to be S. 274 00:19:57,820 --> 00:20:03,080 Because this is a sum I could also think of this as e to the positive 0.015 275 00:20:03,080 --> 00:20:09,180 so that's going to cancel out my e to the negative 0.015 or you could also think 276 00:20:09,180 --> 00:20:14,210 of it as just I'm adding those two numbers together getting e to the 0 and that's 1. 277 00:20:14,210 --> 00:20:18,160 So this bit here is going to cancel out this bit here. 278 00:20:19,620 --> 00:20:25,690 This e to the log of s, that's going to be 279 00:20:25,690 --> 00:20:28,760 just s, so that's going to cancel out my s down here, 280 00:20:31,260 --> 00:20:36,660 so what I am left with is this second term the, the 281 00:20:36,660 --> 00:20:40,279 magnitude of it being equal to 10 times phi of d1 of s. 282 00:20:41,300 --> 00:20:41,300 [INAUDIBLE] 283 00:20:41,300 --> 00:20:41,300 . 284 00:20:41,300 --> 00:20:45,370 So now, let's go back and look at my, my formula for the delta of c. 285 00:20:46,850 --> 00:20:52,400 And so the delta of c was originally phi of d1, 286 00:20:52,400 --> 00:20:58,180 plus 10 times lower-case phi of d1. And then minus. 287 00:20:58,180 --> 00:21:00,446 And we've just showed that this thing, 10 over s. 288 00:21:00,446 --> 00:21:04,010 Times e to the minus 0.01 times phi of d2 of s. 289 00:21:04,010 --> 00:21:06,786 That was the same thing as 290 00:21:06,786 --> 00:21:16,710 10 times phi of d1 of s. So what 291 00:21:16,710 --> 00:21:19,340 I end up with is capital phi of d1 plus 10 lower-case phi minus 10 lower-case phi. 292 00:21:19,340 --> 00:21:22,140 And in the end we find that the delta for 293 00:21:22,140 --> 00:21:27,560 this call option, is actually just capital phi of D 1.