Programming/Kdb/Labs/Option pricing - Revision history

Admin at 22:12, 21 December 2021

2021-12-21T22:12:53Z

← Older revision		Revision as of 22:12, 21 December 2021
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	=Task 1: Implementing the standard normal cumulative distribution function=		=Task 1: Implementing the standard normal cumulative distribution function=

	As mentioned in the ~~<span class="plainlinks">~~[https://www.amazon.co.uk/Handbook-Mathematical-Functions-Formulas-Graphs/dp/161427617X/ Handbook of Mathematical Functions]~~</span>~~, <math>N(x)</math> can be approximated by		As mentioned in the [https://www.amazon.co.uk/Handbook-Mathematical-Functions-Formulas-Graphs/dp/161427617X/ Handbook of Mathematical Functions], <math>N(x)</math> can be approximated by
	<center><math>		<center><math>
	\left\{		\left\{

Admin: /* Task 5: Pricing a double digital option */

2021-06-18T07:47:03Z

Task 5: Pricing a double digital option

← Older revision		Revision as of 07:47, 18 June 2021
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	First, we implement the double digital payoff:		First, we implement the double digital payoff:
	<pre>		<pre>
	double_digital_payoff:{[L;U;S]?[(S<L)\|S>U;~~0fS~~;1f+0fS]}		double_digital_payoff:{[L;U;S]?[(S<L)\|S>U;z;1f+z:0f*S]}
	</pre>		</pre>

Admin: /* Task 4: Implement a Monte Carlo pricer */

2021-06-18T07:43:26Z

Task 4: Implement a Monte Carlo pricer

← Older revision		Revision as of 07:43, 18 June 2021
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	<pre>		<pre>
	mc:{[S;r;q;sigma;T;path_count;payoff]		mc:{[S;r;q;sigma;T;path_count;payoff]
	variance:sigmasigmaT;		root_variance:sqrt variance:sigmasigmaT;
	~~root_variance~~:~~sqrt[variance];~~		moved_spot:Sexp (ito_correction:-.5variance)+(r-q)*T;
	ito_correction:-.5*variance;		exp[neg Tr]avg payoff moved_spotexp[root_variancenormal_variates path_count]};
	~~moved_spot:Sexp ito_correction~~+(r-q)T;
	~~gaussians:normal_variates[path_count];~~
	~~this_spot:moved_spotexp[root_variancegaussians]~~;
	exp[neg Tr]avg payoff ~~this_spot~~};
	</pre>		</pre>

Admin at 00:51, 18 June 2021

2021-06-18T00:51:43Z

← Older revision		Revision as of 00:51, 18 June 2021
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	mc[100f;.05;.07;.1;.5;100000;put_payoff[105f]]		mc[100f;.05;.07;.1;.5;100000;put_payoff[105f]]
	</pre>		</pre>

			=Task 5: Pricing a double digital option=

			We shall now use our q Monte Carlo pricer to price a double digital option.

			First, we implement the double digital payoff:
			<pre>
			double_digital_payoff:{[L;U;S]?[(S<L)\|S>U;0fS;1f+0fS]}
			</pre>

			We are now ready to price a double digital option:
			<pre>
			mc[100f;.05;.07;.1;.5;100000;double_digital_payoff[105f;110f]]
			</pre>

			The result will vary (due to the Monte Carlo randomness), but it will be around 0.1260532.

Admin at 00:39, 18 June 2021

2021-06-18T00:39:41Z

← Older revision		Revision as of 00:39, 18 June 2021
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	<center><math>		<center><math>
	Z_1 = \sqrt{-2 \ln U_1} \sin(2\pi U_2).		Z_1 = \sqrt{-2 \ln U_1} \sin(2\pi U_2).
	</math</center>		</math></center>
	Then <math>Z_0</math> and <math>Z_1</math> are independent random variables with a standard normal distribution.		Then <math>Z_0</math> and <math>Z_1</math> are independent random variables with a standard normal distribution.

Admin at 00:39, 18 June 2021

2021-06-18T00:39:07Z

← Older revision		Revision as of 00:39, 18 June 2021
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	</math></center>		</math></center>
	We do this many times and take the average. We then multiply this average by <math>e^{-rT}</math> and we are done.		We do this many times and take the average. We then multiply this average by <math>e^{-rT}</math> and we are done.

			=Task 3: Generating standard Gaussian random variates=

			To generate 10 standard uniform random variates in q, we can simply use
			<pre>
			10?1f
			</pre>

			How can we convert these standard uniform random variates into standard Gaussian random variates?

			One approach is to use the '''Box-Muller transform'''. Suppose <math>U_1</math> and <math>U_2</math> are independent samples chosen from the uniform distribution on the unit interval <math>[0, 1]</math>. Let
			<center><math>
			Z_0 = \sqrt{-2 \ln U_1} \cos(2\pi U_2)
			</math></center>
			and
			<center><math>
			Z_1 = \sqrt{-2 \ln U_1} \sin(2\pi U_2).
			</math</center>
			Then <math>Z_0</math> and <math>Z_1</math> are independent random variables with a standard normal distribution.

			Equipped with Box-Muller transform, implement a function in q that, given a number <math>x</math>, will return <math>x</math> standard Gaussian random variates.

			<pre>
			normal_variates:{$[x=2n:x div 2;raze sqrt[-2log n?1f]/:(sin;cos)@\:(2pi)*n?1f;-1_.z.s 1+x]}
			</pre>

			=Task 4: Implement a Monte Carlo pricer=

			We are now ready to implement a Monte Carlo pricer. We can use it to price more complicated options than the European call and put. However, the European call and put are convenient — the analytic solutions enable us to test our Monte Carlo pricer.

			<pre>
			mc:{[S;r;q;sigma;T;path_count;payoff]
			variance:sigmasigmaT;
			root_variance:sqrt[variance];
			ito_correction:-.5*variance;
			moved_spot:Sexp ito_correction+(r-q)T;
			gaussians:normal_variates[path_count];
			this_spot:moved_spotexp[root_variancegaussians];
			exp[neg Tr]avg payoff this_spot};
			</pre>

			Let's define the European call and put payoffs:

			<pre>
			call_payoff:{[K;S]0\|S-K}
			put_payoff:{[K;S]0\|K-S}
			</pre>

			We can now test our Monte Carlo pricer on these payoffs and confirm that we get similar answers to those produced by the analytic formulae:

			<pre>
			mc[100f;.05;.07;.1;.5;100000;call_payoff[105f]]
			mc[100f;.05;.07;.1;.5;100000;put_payoff[105f]]
			</pre>

Admin at 00:29, 18 June 2021

2021-06-18T00:29:25Z

@@ Line 146: / Line 146: @@
 The price of a vanilla option is therefore equal to
 <center><math>
-e^{-rT} \mathbb{E}\left[ S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right] \right].
+e^{-rT} \mathbb{E}\left[ f\left( S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right] \right) \right].
 </math></center>

Admin at 00:25, 18 June 2021

2021-06-18T00:25:15Z

← Older revision		Revision as of 00:25, 18 June 2021
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	<center><math>		<center><math>
	S_T = S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right].		S_T = S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right].
			</math></center>

			The price of a vanilla option is therefore equal to
			<center><math>
			e^{-rT} \mathbb{E}\left[ S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right] \right].
	</math></center>		</math></center>

Admin at 00:24, 18 June 2021

2021-06-18T00:24:02Z

← Older revision		Revision as of 00:24, 18 June 2021
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	<center><math>		<center><math>
	\ln S_T = \ln S_0 + \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1),		\ln S_T = \ln S_0 + \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1),
			</math></center>
			or equivalently,
			<center><math>
			S_T = S_0 \exp \left[ \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1) \right].
	</math></center>		</math></center>

Admin at 00:23, 18 June 2021

2021-06-18T00:23:09Z

← Older revision		Revision as of 00:23, 18 June 2021
Line 138:		Line 138:
	<center><math>		<center><math>
	\ln S_T = \ln S_0 + \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1),		\ln S_T = \ln S_0 + \left(r - q - \frac{q}{2}\sigma^2\right) T + \sigma \sqrt{T} N(0, 1),
	</~~center~~></~~math~~>		</math></center>