Added CLT section

Author: dingran

README.md

@@ -1,4 +1,4 @@
-# Quantitative interview questions and strategies
+# Quantitative Finance Interview Questions and Strategies
 
 ## Introduction
 
@@ -92,7 +92,7 @@ Probability problems should be fun to solve and let's begin!
     - simplex
     - frog jump
 
-- Approximation trick1: Central limit theorem
+- Approximation trick1: [Central limit theorem](prob_clt.ipynb)
     - fake coin
     - monte carlo integration
 

prob_clt.ipynb

@@ -0,0 +1,93 @@
+{
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+   "metadata": {},
+   "source": [
+    "# Central Limit Theorem (CLT)\n",
+    "\n",
+    "## Definition:\n",
+    "Let $X_{1}$, $X_{2}$, $X_{3}$,... be i.i.d with mean $\\mu$ and variance $\\sigma^{2}$. As $n \\rightarrow \\infty$, let $S=\\sum_{k=1}^n X_{i}$, we have $S \\rightarrow \\mathcal{N}(n\\mu, n\\sigma^{2})$ and $\\frac{S-n\\mu}{\\sqrt{n\\sigma^{2}}} \\rightarrow \\mathcal{N}(0,1)$\n",
+    "\n",
+    "Equivalently, let $M=\\frac{1}{n}\\sum_{k=1}^n X_{i}$, we have\n",
+    "$M \\rightarrow \\mathcal{N}(\\mu,\\sqrt{\\frac{\\sigma^2}{n}})$ and $\\frac{M-\\mu}{\\sqrt{\\frac{\\sigma^2}{n}}} \\rightarrow \\mathcal{N}(0,1)$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Discussions:\n",
+    "\n",
+    "Naturally CLT appears in questions that invovles sum or average of a large number of random variable and especially when the question only asks for an approximate answer. Here are a few quick examples."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "<br>\n",
+    "***Example1:***\n",
+    "*Supposed we have a fair coin and we flip it 400 times. What is the probability you will see 210 heads or more.*\n",
+    "\n",
+    "<br>\n",
+    "**Exact answer**\n",
+    "\n",
+    "Let the outcome of each coin flip be a random variable $I_{i}$. Thus we are dealing with the random variable $S=\\sum_{i=1}^{400}I_{i}$. $S$ is te sume of a series of i.i.d Bernoulie trials, thus it follows Binomial distribution. So the exact answer is: $P(S\\geq210)= \\sum_{k=210}^{400}C_{400}^{k}\\left(\\frac{1}{2}\\right)^{400}$ which requires a program to calculate (Actually try implementing this, beware of roudoff errors and compare it against the approximate answer below.).\n",
+    "\n",
+    "<br>\n",
+    "**Approximation**\n",
+    "\n",
+    "We use CLT to easily get an approxmate answer quickly. First recognize that for each $I_{i}$ we have $\\mu=0.5$ and $\\sigma^2=0.5\\times(1-0.5)=0.25$. Then, $Z=\\frac{S-400*0.5}{\\sqrt{400*0.25}}=\\frac{S-200}{10}$ is approximately $\\mathcal{N}(0,1)$. For $S \\geq 210$, we have $Z\\geq1$. The 68-95-99.7 rule tells us that for a standardized normal distribution, the probability of the random variable taking value more than 1 standard deviation away from the center is $1-0.68=0.32$ and thus the one sided probability for $P(Z\\geq1) = 0.16$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "<br>\n",
+    "***Example2:***\n",
+    "\n",
+    "*Supposed you are going to use Monte Carlo simulation to estimate value of $\\pi$. How would you implement it? If we require an error of 0.001, how many trials/ data points do you need?*\n",
+    "\n",
+    "**Solution**\n",
+    "\n",
+    "One possible implementation is to have a rectangle, say $x \\in [-1,1], y\\in[-1,1]$. If we uniformly randomly draw a point from this rectangle, the probability of the point following into the circle region $x^2+y^2\\lt1$ is the ratio of the area between the circle and rectangle. \n",
+    "\n",
+    "Formally, let random indicator variable $I$ take value 1 if the point falls in the circle and 0 otherwise, then $p=P(I=1)=\\frac{\\pi}{4}$ and $E(I)=p$. If we do $n$ such trials, and define $M=\\frac{1}{n}\\sum_{k=1}^n I_{i}$, then $M$ follows approximately $\\mathcal{N}(\\mu_{I},\\frac{\\sigma_{I}^2}{n})$. In this setup, $\\mu_{I}=p=\\frac{\\pi}{4}$ and $\\sigma_{I}^2=p(1-p)$.\n",
+    "\n",
+    "One thing we need to clarify with the interviewer is what error really means? She might tell you just consider this it as the standard deviation of your estimated $\\pi$. Therefore the specified error translates into a required sigma of $\\sigma_{req}=\\frac{error}{4}$ for random variable $M$. Thus $n = \\frac{\\sigma_{I}^2}{\\sigma_{req}^2}$, it is about 2.7 million for our particular case.\n",
+    "\n",
+    "We can see that the number of trials $n$ scales with $\\frac{1}{error^2}$, which is caused by the $\\frac{1}{\\sqrt{n}}$ scaling of the $\\sigma_{M}$ in the CLT, and is generally the computation complexity in Monte Carlo integration.\n"
+   ]
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