1 00:00:01,040 --> 00:00:03,460 So moving on for the next lesson. 2 00:00:03,460 --> 00:00:08,800 We'll talk about gamma, which is the derivative of delta with respect 3 00:00:08,800 --> 00:00:12,860 to s, or the second derivative of c with respect to s. 4 00:00:15,540 --> 00:00:18,020 So gamma is the rate of change of delta. 5 00:00:20,350 --> 00:00:23,720 So I can write that as the partial derivative with respect to s of delta. 6 00:00:24,810 --> 00:00:28,710 And delta is the partial derivative of c with respect to s. 7 00:00:28,710 --> 00:00:32,450 So this is just the second derivative of c with respect to s. 8 00:00:33,460 --> 00:00:35,092 But again, the way I'm going to compute 9 00:00:35,092 --> 00:00:37,680 it, since I've already computed this derivative for delta. 10 00:00:38,720 --> 00:00:42,160 I'm going to use that as a starting point rather than starting from scratch. 11 00:00:44,730 --> 00:00:48,080 So all we have to do, since we already have an expression for 12 00:00:48,080 --> 00:00:51,940 delta, is just take the derivative with respect to s of that again. 13 00:00:53,300 --> 00:00:57,404 But that turns out to be pretty difficult, because in the last in 14 00:00:57,404 --> 00:01:02,600 the last set of slides, we, we had this three term expression for s. 15 00:01:02,600 --> 00:01:04,080 But luckily, there's a shortcut. 16 00:01:04,080 --> 00:01:07,104 So like I mentioned in the previous set, if 17 00:01:07,104 --> 00:01:10,128 we look at the second two terms in the square 18 00:01:10,128 --> 00:01:14,790 brackets here, I said that those were going to be equal to zero. 19 00:01:14,790 --> 00:01:18,370 But that's something that we have to show mathematically before we can do it. 20 00:01:18,370 --> 00:01:19,321 We did this. 21 00:01:19,321 --> 00:01:21,369 I had one example of this before where I 22 00:01:21,369 --> 00:01:24,825 had actually plugged in numbers for the Black-Scholes pricing 23 00:01:24,825 --> 00:01:27,321 formula, and we went through the, the work of 24 00:01:27,321 --> 00:01:30,082 showing that this part ends up being equal to zero. 25 00:01:30,082 --> 00:01:35,162 And now we'll do it in general so we never have to 26 00:01:35,162 --> 00:01:36,603 do it again. 27 00:01:36,603 --> 00:01:39,879 So the textbook says that delta is just equal to 28 00:01:39,879 --> 00:01:44,219 this much simpler expression than what we've come up with. 29 00:01:45,950 --> 00:01:48,984 And finding an expression for gamma is certainly 30 00:01:48,984 --> 00:01:50,760 going to be a lot easier if I only 31 00:01:50,760 --> 00:01:53,498 have to take the derivative of this, rather 32 00:01:53,498 --> 00:01:56,540 than taking the derivative of all three terms. 33 00:01:58,090 --> 00:02:00,300 So what I'd like to do is figure 34 00:02:00,300 --> 00:02:04,650 out a way to show that this expression here is equal to zero. 35 00:02:04,650 --> 00:02:08,290 And so when I, when I write something like this, I'm not sure 36 00:02:08,290 --> 00:02:12,170 that it's equal to zero, I want to show that it's equal to zero. 37 00:02:12,170 --> 00:02:14,712 So you don't want to just write down equal to zero 38 00:02:14,712 --> 00:02:18,740 and then start plugging in things and showing that it's true. 39 00:02:18,740 --> 00:02:24,570 What I want to do is start with this without saying it's equal to zero. 40 00:02:24,570 --> 00:02:26,010 And just do algebraic 41 00:02:26,010 --> 00:02:29,930 manipulations to it, until it finally is equal to zero. 42 00:02:29,930 --> 00:02:33,680 And then what we'll have is, a line of work that starts 43 00:02:33,680 --> 00:02:38,160 out from what I'm given, and ends with what I want, basically zero. 44 00:02:39,790 --> 00:02:43,580 And that'll be an argument that anytime I see this. 45 00:02:43,580 --> 00:02:48,370 Expression in a calculation, I can then replace that with zero. 46 00:02:48,370 --> 00:02:51,139 And that's going to save us a lot of work in, 47 00:02:51,139 --> 00:02:53,103 the next several lessons. 48 00:02:55,210 --> 00:03:00,670 And so the strategy I'm going to use is I have a, one term that has 49 00:03:00,670 --> 00:03:04,910 a d plus in it and one term that has a d minus in it. 50 00:03:06,490 --> 00:03:09,690 But remember d minus could be written in terms of 51 00:03:09,690 --> 00:03:12,960 d plus just by adding a little bit to it. 52 00:03:12,960 --> 00:03:16,527 And so what I'm going to do is try and manipulate the 53 00:03:16,527 --> 00:03:20,442 d minus into a d plus, and then there'll be some stuff 54 00:03:20,442 --> 00:03:21,820 left over. 55 00:03:21,820 --> 00:03:26,740 And hopefully when I factor out the stuff that's left over, it's going to. 56 00:03:26,740 --> 00:03:30,820 Turn this second term into the first term, and if both of those terms are 57 00:03:30,820 --> 00:03:35,160 the same, then when I subtract one from the other, I'm going to get zero. 58 00:03:37,950 --> 00:03:45,470 So phi of d minus is equal to 1 over the square root of 2 pi e. 59 00:03:45,470 --> 00:03:48,275 So I'm using some different notation here just because it, 60 00:03:48,275 --> 00:03:51,110 it gets really small if I have everything in the exponent. 61 00:03:51,110 --> 00:03:56,150 So, exp as a function, so exp and then something in parentheses that means 62 00:03:56,150 --> 00:04:01,600 exactly the same thing as e to the power of what's ever in the parentheses here. 63 00:04:04,220 --> 00:04:09,575 So this is 1 over the square root of 2 pi Square root of 2 pi times 64 00:04:09,575 --> 00:04:14,939 e to the minus d minus squared over 2. And that's just the, 65 00:04:14,939 --> 00:04:19,886 the exact definition of phi of d minus. And 66 00:04:19,886 --> 00:04:24,926 now I'm going to replace d minus in that expression, 67 00:04:24,926 --> 00:04:29,270 with d plus minus. Sigma times the square root of t minus t. 68 00:04:29,270 --> 00:04:31,600 And that was just my expression for d minus. 69 00:04:33,850 --> 00:04:38,780 And now I have a, this quadratic, form here. 70 00:04:38,780 --> 00:04:40,400 So I'm going to expand that. 71 00:04:41,610 --> 00:04:45,320 The, the first term, I'm going to end up with a d plus squared over 2. 72 00:04:45,320 --> 00:04:49,840 So it's pretty clear that that's going to, that's going to get me a phi of d plus. 73 00:04:49,840 --> 00:04:53,870 And then hopefully the stuff that's left over is going to be useful to get this 74 00:04:53,870 --> 00:04:56,412 result that I'm after, showing that the second 75 00:04:56,412 --> 00:04:58,670 term here is actually equal to the first. 76 00:05:02,630 --> 00:05:06,954 So this is just what I had on the last bit of the, of the previous slide. 77 00:05:08,480 --> 00:05:13,838 So now what I'm going to do is just go ahead and expand this quadratic 78 00:05:13,838 --> 00:05:19,290 form here, and so I'll end up with d plus squared minus 2d plus sigma 79 00:05:19,290 --> 00:05:24,830 square root of t minus t plus sigma squared times t minus t all over 2. 80 00:05:24,830 --> 00:05:28,834 And now I'm going to regroup 81 00:05:28,834 --> 00:05:32,120 the terms like this. 82 00:05:32,120 --> 00:05:35,340 So basically, I'm just taking, ooh, got the whole thing that time. 83 00:05:37,040 --> 00:05:42,590 I'm just taking this d plus squared over 2 and putting that by itself. 84 00:05:42,590 --> 00:05:45,762 Because this is now an expression, the 1 over the square root 85 00:05:45,762 --> 00:05:49,410 of 2 pi E to the minus d plus squared over two. 86 00:05:49,410 --> 00:05:52,034 That's just and expression from phi of d plus. 87 00:05:52,034 --> 00:05:54,254 And then these two terms 88 00:05:54,254 --> 00:05:59,471 are what's left over. So the middle term that has 89 00:05:59,471 --> 00:06:04,736 a two in the numerator and in the denominator, 90 00:06:04,736 --> 00:06:09,770 so those are going to cancel each other out. 91 00:06:11,342 --> 00:06:13,940 So I've, I've got from my, oh I can't highlight anything. 92 00:06:15,410 --> 00:06:20,328 I have my phi of d plus from this portion of the previous line. 93 00:06:20,328 --> 00:06:27,160 I have e to the d plus times sigma square root of t minus t from this portion. 94 00:06:27,160 --> 00:06:28,909 And then I have e, this hasn't changed 95 00:06:28,909 --> 00:06:31,130 I've just written it a little bit differently. 96 00:06:32,190 --> 00:06:35,670 e to the minus sigma squared t minus t divided by 2. 97 00:06:35,670 --> 00:06:44,970 And now, just a quick reminder, of what d plus is equal to so what I'm going to. 98 00:06:44,970 --> 00:06:45,330 What 99 00:06:45,330 --> 00:06:48,750 I have here is d plus times sigma square root of 100 00:06:48,750 --> 00:06:51,990 t minus t and so all these sigma square root t 101 00:06:51,990 --> 00:06:55,230 minus t is going to do is cancel out the sigma 102 00:06:55,230 --> 00:07:00,010 square root of t minus t in the denominator of d plus. 103 00:07:00,010 --> 00:07:05,680 So I, what I have is E to the numberator of the definition of d plus here. 104 00:07:05,680 --> 00:07:10,590 So, I'll do that in the next line. So, all that's 105 00:07:10,590 --> 00:07:15,650 happened is this d plus sigma square root of t minus t has 106 00:07:15,650 --> 00:07:20,750 just become the numerator of the definition of d plus. 107 00:07:22,460 --> 00:07:24,924 And then the bit on the right hand term is still stayed 108 00:07:24,924 --> 00:07:28,249 the same, just sigma squared, t minus t over 2 and it's negative. 109 00:07:35,830 --> 00:07:41,620 So, what's going to happen, I have e to the log of s over k. 110 00:07:41,620 --> 00:07:43,943 So the exponential and the logarithm, they are. 111 00:07:44,960 --> 00:07:47,392 Inverse functions of one another it's the, 112 00:07:47,392 --> 00:07:50,016 the exponential undoes what the logarithm did, 113 00:07:50,016 --> 00:07:52,880 so this is just going to allow me to take an s over k out. 114 00:07:54,630 --> 00:07:57,754 And then when I have an exponential of a sum, that's 115 00:07:57,754 --> 00:08:00,878 equal to the, product, so this is the same thing as e 116 00:08:00,878 --> 00:08:02,130 to the log. 117 00:08:02,130 --> 00:08:08,190 S over k times e to the r minus q plus sigma squared over 2, t minus t. 118 00:08:11,040 --> 00:08:16,528 So this bit here is going to stay in my exponential function 119 00:08:16,528 --> 00:08:21,792 but I'm going to factor out an s over k and then also using 120 00:08:21,792 --> 00:08:27,690 the same property So here I had a product of exponentials. 121 00:08:27,690 --> 00:08:30,657 So this is just going to be the sum of this bit that I've 122 00:08:30,657 --> 00:08:35,360 got highlighted and what's inside the square brackets in the lower right here. 123 00:08:37,230 --> 00:08:40,428 So that gives this s over k out of my exponential 124 00:08:40,428 --> 00:08:46,000 bit, and then leaves me with what's inside the exponential bit here. 125 00:08:46,000 --> 00:08:49,510 But now look what's happened, so this, sigma squared t minus t 126 00:08:49,510 --> 00:08:53,580 over two that I've been carrying down the right side of my calculation. 127 00:08:54,670 --> 00:08:57,628 I have a plus sigma squared over 2 times t 128 00:08:57,628 --> 00:09:02,040 minus t minus sigma squared over 2 times t minus t. 129 00:09:02,040 --> 00:09:02,330 So all 130 00:09:02,330 --> 00:09:05,530 of that's going to cancel out, and I'm just going to have r minus q. 131 00:09:05,530 --> 00:09:08,770 That quantity times t minus t left. 132 00:09:11,150 --> 00:09:12,650 And so now I'm going to go back to 133 00:09:12,650 --> 00:09:16,100 using the notation of writing this as exponential. 134 00:09:16,100 --> 00:09:21,410 So, this is e to the r times t minus t. 135 00:09:21,410 --> 00:09:22,740 So that's where this bit came from. 136 00:09:22,740 --> 00:09:29,435 And then this is e to the minus q. Times t minus t, so that's 137 00:09:29,435 --> 00:09:36,335 where this bit came from and now I have a, an expression for phi of d minus 138 00:09:36,335 --> 00:09:39,900 in terms of phi of d plus and this is the 139 00:09:39,900 --> 00:09:43,465 expression that I am trying to show is 140 00:09:43,465 --> 00:09:47,293 equal to zero again. So I am going to substitute. 141 00:09:47,293 --> 00:09:54,390 My expression here for phi of d minus into the second term 142 00:09:54,390 --> 00:10:01,356 here and I get this expression here. And so the 143 00:10:01,356 --> 00:10:06,839 first term has stayed the same but now I have k times e to the minus r t minus t. 144 00:10:08,770 --> 00:10:13,799 So that carried over from before, and now I have what's 145 00:10:13,799 --> 00:10:18,828 coming from my new expression, fee of d plus, s divided 146 00:10:18,828 --> 00:10:23,650 by k, e to the rt minus t, and e to the minus qt minus t. 147 00:10:24,690 --> 00:10:27,040 But here I have a k And 148 00:10:27,040 --> 00:10:30,550 s over k, so that's going to be just s, and I have 149 00:10:30,550 --> 00:10:34,800 an s in the numerator, and an s in the denominator here. 150 00:10:34,800 --> 00:10:37,440 So all of the strike prices, and the, the asset 151 00:10:37,440 --> 00:10:40,350 prices are going to cancel each other out in this term. 152 00:10:41,940 --> 00:10:47,220 Then I have e to the minus rt minus t, and 153 00:10:47,220 --> 00:10:52,086 e to the rt minus t. So when I add those together I have 154 00:10:52,086 --> 00:10:54,320 minus rt minus t plus rt minus t. 155 00:10:54,320 --> 00:10:58,230 That's the same things as e to the zero which is just one. 156 00:10:58,230 --> 00:11:00,490 So, those two parts are going to cancel each other out. 157 00:11:03,070 --> 00:11:10,910 So, all I'm left with Is phi of d plus. And e to the minus q t minus t. 158 00:11:12,190 --> 00:11:14,580 Which is exactly what I had in the numerator over here. 159 00:11:16,260 --> 00:11:21,500 And now in the denominator, remember I've canceled out this s here with this s here. 160 00:11:21,500 --> 00:11:23,280 So I have the same denominator as well. 161 00:11:26,360 --> 00:11:28,536 So when I do all of the canceling here, 162 00:11:28,536 --> 00:11:31,810 I've made the second term into the first term. 163 00:11:31,810 --> 00:11:35,160 And I'm subtracting them from one another, so I end up having zero. 164 00:11:35,160 --> 00:11:40,520 So that means when it's time to actually go ahead and calculate gamma. 165 00:11:41,820 --> 00:11:47,058 Instead of having to take the derivative with respect to s of this 166 00:11:47,058 --> 00:11:52,005 great big formula, I can rewrite that as just the derivative 167 00:11:52,005 --> 00:11:57,879 with respect to s of e to the minus q t minus t times capital phi of d plus. 168 00:12:02,470 --> 00:12:06,270 And so when I'm taking the partial derivative with respect 169 00:12:06,270 --> 00:12:08,320 to S, e to the minus q t minus t. 170 00:12:08,320 --> 00:12:10,950 There's no S in that, so I can treat that as a constant. 171 00:12:10,950 --> 00:12:13,990 So I can move that across the derivative operator. 172 00:12:13,990 --> 00:12:15,784 And so all I need to do is take the 173 00:12:15,784 --> 00:12:18,899 derivative with respect to s of capital phi of d plus. 174 00:12:21,440 --> 00:12:23,288 And I'll go through this again, but we 175 00:12:23,288 --> 00:12:25,980 actually did this already when we were computing delta. 176 00:12:27,240 --> 00:12:30,255 So, I'm taking the derivative of a function of d plus 177 00:12:30,255 --> 00:12:33,770 with respect to s so I have to use the chain rule. 178 00:12:33,770 --> 00:12:37,700 And that tells me that I can take the derivative of capital phi of d plus. 179 00:12:37,700 --> 00:12:43,530 And then I have to multiply that by the derivative of d plus with respect to s. 180 00:12:47,150 --> 00:12:49,880 And we've all ready calculated that as well so. 181 00:12:51,212 --> 00:12:55,440 partial derivative of either d plus or d minus with respect to s. 182 00:12:55,440 --> 00:12:58,860 It's just one over s times sigma times the square root of t minus t 183 00:13:01,360 --> 00:13:05,581 And then this we've already seen before let's say this is the derivative of our 184 00:13:05,581 --> 00:13:07,912 indefinite integral so this is just going to to 185 00:13:07,912 --> 00:13:09,855 be equal to lowercase phi of d plus. 186 00:13:13,380 --> 00:13:19,340 So what I'm left with is that e to the minus qt minus t that I had out here. 187 00:13:20,850 --> 00:13:26,240 The 1 over s times sigma times the square root of t minus t that came 188 00:13:26,240 --> 00:13:31,900 from the chain rule so that's in the denominator in the front here. 189 00:13:31,900 --> 00:13:34,700 And, then, the lower case fee of d plus, 190 00:13:34,700 --> 00:13:37,690 which was just a derivative that I got here. 191 00:13:41,980 --> 00:13:46,050 And so, we now have an expression for gamma, which is the partial 192 00:13:46,050 --> 00:13:51,230 derivative with respect to S of, sorry, of delta with respect to S. 193 00:13:51,230 --> 00:13:53,690 And this is exactly the same as the line above it. 194 00:13:53,690 --> 00:13:57,810 I've just replaced phi of d plus with the actual definition of phi of d plus. 195 00:13:57,810 --> 00:14:04,040 So, it's 1 over the square root of 2 pi. E to the minus d plus squared over 2, so. 196 00:14:04,040 --> 00:14:07,029 Looks like I missed a little bit because I forgot the square 197 00:14:07,029 --> 00:14:09,667 there, but this should be d plus squared over 2.