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Archive for August 2010

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Written by curious

August 23, 2010 at 9:32 am

Posted in quant-finance

QUANT: Structured Product – Accumulator

Accumulator (or Share Forward Accumulators) require the issuer to sell shares of some underlying instrument at a predetermined strike price, settled periodically. The investor accumulates the underlying stock holdings over the term of the contract.

Bushels are priced above today’s current market & there is no up front premium costs. There is potential of weekly commitment doubling in the contract period. If the knockout level is hit, unaccumulated bushels are unpriced.

Written by curious

August 16, 2010 at 10:56 am

Posted in quant-finance


  1. Open VB Editor: Alt+F11. Or Developer->Visual Basic. If Developer menu item is not visible, then OfficeButton->Excel Options->Show Developer tab in the Ribbon.
  2. How do I view a list of all Cell Reference names in a spreadsheet? Use Formulas->Name Manager (Ctrl+F3).
  3. How can I create a drop down list for a cell? Use Data Validation. Enter data in a column in your spreadsheet. Then for a cell define Data Validation to allow “List” and set the “Source” to the cells containing items for drop down list.

Written by curious

August 12, 2010 at 10:10 am

Posted in tech-tips

STAT: Multivariate

Distribution function for random variable \mathbf{X} = (X_1, X_2, \ldots, X_n)

    F_{\mathbf{X}}(x_1, x_2, \ldots, x_n) = P(X_1 \leq x_1, X_2 \leq x_2, \ldots, X_n \leq x_n)
    F_{\mathbf{X}}(x_1,\ldots,x_d,\infty\ldots,\infty) = F_{X_1,\ldots,X_d}(x_1,\ldots,x_d)
    F_{\mathbf{X}}(x_1,\ldots,x_d,\ldots,-\infty) = 0

Density Function

    f(x_1,x_2,\ldots,x_n) \equiv \frac{\partial^n}{\partial x_1 \partial x_2 \ldots \partial x_n} F(x_1,x_2,\ldots,x_n)
    f_{X_1,X_2,\ldots,X_k}(x_1,x_2,\ldots,x_k) = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(x_1,x_2,\ldots,x_n) dx_{k+1}\ldots dx_n


    E[g(X_1,\ldots,X_n)] = \int_{-\infty}^{\infty} \ldots \int_{-\infty}^{\infty} g(x_1,\ldots,x_n) f(x_1,\ldots,x_n) dx_1 \ldots dx_n


  • P(X_1 \in A_1, \ldots, X_n \in A_n) = P(X_1 \in A_1) \ldots P(X_n \in A_n)
  • F(x_1,x_2,\ldots,x_n) = \prod_i F_{X_i}(x_i)
  • f(x_1,x_2,\ldots,x_n) = \prod_i f_{X_i}(x_i)
  • E[\prod_i g_i(X_i)] = \prod_i E[g_i(X_i)]
  • Given independent Xi‘s such that Y=\sum X_i, \quad M_Y[t] = \prod_i M_{X_i}[t]

Multivariate Moment Generating Function

    M_{\mathbf{X}}(\mathbf{t}) = E[e^{\mathbf{t}'\mathbf{X}}] = E[e^{\sum X_i t_i}]

Independent Xi:

    M_{\mathbf{X}}(\mathbf{t}) = \prod M_{X_i}(t_i)

Conditional Density
Let \mathbf{X} = (\mathbf{X_1} : \mathbf{X_2}), \mathbf{X_1} = (x_1,\ldots,x_k), \mathbf{X_2} = (x_{k+1},\ldots,x_n)

    f_{\mathbf{X_1}|\mathbf{X_2}}(\mathbf{x}_1|\mathbf{x}_2) = \frac{f(\mathbf{x}_1,\mathbf{x}_2)}{f_{\mathbf{X}_2}(\mathbf{x}_2)}

Covariance Matrix

  • \boldsymbol{\Sigma}_{\mathbf{X}} = \text{Cov}(\mathbf{X},\mathbf{X}) = E[(\mathbf{X}-\boldsymbol{\mu})(\mathbf{X}-\boldsymbol{\mu})^T], \quad \text{ where } \boldsymbol{\mu} = E[\mathbf{X}]
  • \boldsymbol{\Sigma}_{\mathbf{X}} = \text{Cov}(\mathbf{X},\mathbf{X}) = E[\mathbf{X}\mathbf{X}^T]-\boldsymbol{\mu}\boldsymbol{\mu}^T
  • \text{Cov}(\mathbf{X},\mathbf{X}) = E[\text{Cov}(\mathbf{X},\mathbf{X})|\mathbf{Y}] + \text{Cov}(E[\mathbf{X}|\mathbf{Y}],E[\mathbf{X}|\mathbf{Y}])
  • Let \mathbf{Y} = \mathbf{a}+\mathbf{B}\mathbf{X}, then \boldsymbol{\mu}_{\mathbf{Y}} = \mathbf{a}+\mathbf{B}\boldsymbol{\mu}_{\mathbf{X}} and \text{Cov}(\mathbf{Y},\mathbf{Y}) = \boldsymbol{\Sigma}_{\mathbf{Y}} = \mathbf{B} \boldsymbol{\Sigma}_{\mathbf{X}} \mathbf{B}^T
  • Non-negative Definite, i.e., \mathbf{b}^T \boldsymbol{\Sigma}_{\mathbf{X}} \mathbf{b} \geq 0, \quad \forall \mathbf{b} \in \mathbb{R}^n
    Let Y = \mathbf{b}^T \mathbf{X} = \sum b_i X_i, then 0 \leq \text{Var}(Y) = \mathbf{b}^T \boldsymbol{\Sigma}_{\mathbf{X}} \mathbf{b}
  • \mathbf{b} = \mathbf{1} = (1,\ldots,1)^T, Y = \mathbf{1}^T \mathbf{X} = \sum X_i then \text{Var}[\sum X_i] = \mathbf{1}^T \boldsymbol{\Sigma}_{\mathbf{X}} \mathbf{1} = \sum \text{Var}[X_i] + 2 \sum \sum_{i<j} \text{Cov}(X_i,X_j)

Standardization of random variable
Since \boldsymbol{\Sigma}_{\mathbf{X}} is non-negative definite, there exists a matrix A such that

    \mathbf{A}\mathbf{A}^T = \boldsymbol{\Sigma}_{\mathbf{X}}

In addition, if it is positive definite, then A is invertible. Now let

    \mathbf{Z} = \mathbf{A}^{-1}(\mathbf{X}-\boldsymbol{\mu}_{\mathbf{X}})


  • \boldsymbol{\mu}_{\mathbf{Z}} = 0
  • \boldsymbol{\Sigma}_{\mathbf{Z}} = \mathbf{I}
  • Z is standardized: a) Zi are uncorrelated, b) E[Zi] = 0; c) Var[Zi] = 1

Another Standardization of random variable
Let us say

    \mathbf{Z} = \mathbf{D}^{-1}(\mathbf{X}-\boldsymbol{\mu}_{\mathbf{X}})
    where \mathbf{D} = \text{diagonal}(D_{ii}) = \text{diagonal}(\sqrt{\Sigma_{ii}}) = \text{diagonal}(\sigma_i)

i.e. Z_i = \frac{X_i-\mu_{X_i}}{\sigma_i}

Here too Z is standardized: a) but Zi are not uncorrelated, b) E[Zi] = 0; c) Var[Zi] = 1

Actually, covariance matrix of Z is same as correlation matrix of X :- \boldsymbol{\rho}_{\mathbf{X}}

  • \boldsymbol{\Sigma}_{\mathbf{Z}} = \mathbf{D}^{-1} \boldsymbol{\Sigma}_{\mathbf{X}}(\mathbf{D}^{-1})^T = \boldsymbol{\rho}_{\mathbf{X}} \neq \text{diagonal}
  • (\boldsymbol{\Sigma}_{\mathbf{Z}})_{ij} = \sum_k \sum_l (D^{-1})_{ik} \sigma_{kl} (D^{-1})_{lj} = \frac{\sigma_{ij}}{\sigma_{ii}\sigma_{jj}} = \rho_{ij}

Change of variable – multivariate

  1. C be the smallest open set in \mathbb{R}^d so that P(\mathbf{X} \in C) = 1
  2. Let \mathbf{Y} = \mathbf{y}(\mathbf{X}), \quad \mathbf{y} \colon C \to \mathbb{R}^d is one-to-one and continuously differentiable
  3. y has an inverse x. So that X = x(Y)
  4. X has probability density function fX

Then probability density function of Y is given by:

    f_{\mathbf{Y}}(\mathbf{y}) = f_{\mathbf{X}}(\mathbf{x}(\mathbf{y})) \vert \det\left(\frac{\partial \mathbf{x}}{\partial \mathbf{y}}\right) \vert

    Here \left(\frac{\partial \mathbf{x}}{\partial \mathbf{y}}\right) is the Jacobian Matrix defined by
    \left(\frac{\partial \mathbf{x}}{\partial \mathbf{y}}\right)_{ij} = \frac{\partial x_i}{\partial y_j}

Written by curious

August 11, 2010 at 9:37 pm

Posted in statistics

STAT: Bivariate

Distribution function for random variable \mathbf{X} = (X_1, X_2)

    F_{\mathbf{X}}(x_1,x_2) = P(X_1 \leq x_1, X_2 \leq x_2)
    F_{\mathbf{X}}(x_1,\infty) = F_{X_1}(x_1), \quad F_{\mathbf{X}}(\infty,x_2) = F_{X_2}(x_2)
    F_{\mathbf{X}}(x_1,-\infty) = F_{\mathbf{X}}(-\infty,x_2) = 0

For a_1 \leq b_1, a_2 \leq b_2

    P(a_1 < X_1 \leq b_1, a_2 < X_2 \leq b_2) = F(b_1,b_2) + F(a_1,a_2) - F(a_1,b_2) - F(b_1,a_2)

Density Function

    f(x_1,x_2) \equiv \frac{\partial^2}{\partial x_1 \partial x_2} F(x_1,x_2)

    \displaystyle \lim{h \to 0} \frac{P(x_1 < X_1 \leq x_1+h,x_2 < X_2 \leq x_2+h)}{h^2} = f(x_1,x_2)

Marginal Density

    f_{X_1}(x_1) = \int_{-\infty}^{\infty} f(x_1,x_2) dx_2 ; \quad f_{X_2}(x_2) = \int_{-\infty}^{\infty} f(x_1,x_2) dx_1


    E[g(X_1,X_2)] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x_1,x_2) f(x_1,x_2) dx_1 dx_2


    \text{Cov}(X_1,X_2) = E[(X_1-\mu_1)(X_2-\mu_2)] = E[X_1 X_2] - \mu_1 \mu_2
    \text{Cov}(a_1+b_1 X_1, a_2+b_2 X_2) = b_1 b_2 \text{Cov}(X_1,X_2)


    \text{Var}(X_1+X_2) = \text{Var}(X_1) + \text{Var}(X_2) + \text{Cov}(X_1,X_2)

Pearson Correlation

    \rho(X_1, X_2) = \frac{\text{Cov}(X_1,X_2)}{\sqrt{\text{Var}[X_1]\text{Var}[X_2]}}
    \rho(a_1+b_1 X_1, a_2+b_2 X_2) = \text{sign}(b_1 b_2) \rho(X_1,X_2)
    \text{sign}(b) = \begin{cases} 1, & b > 0, \\ 0, & b = 0 \\ -1, & b < 0 \end{cases}


  • P(X_1 \in A_1, X_2 \in A_2) = P(X_1 \in A_1) P(X_2 \in A_2)
  • F(x_1,x_2) = F_{X_1}(x_1)F_{X_2}(x_2), \quad \forall x_1,x_2
  • f(x_1,x_2) = f_{X_1}(x_1)f_{X_2}(x_2), \quad \forall x_1,x_2
  • f(x_1,x_2)dx_1 dx_2 = f_{X_1}(x_1) dx_1 f_{X_2}(x_2) dx_2
  • E[g_1(X_1)g_2(X_2)] = E[g_1(X_1)] E[g_2(X_2)]
    Show that
    1) given this condition implies independence (P(AB) = P(A)P(B)). Use E[1A] = P(A)
    2) given independence, this equality follows. Just follows from definition of E.

X1 and X2 are uncorrelated if Cov(X1, X2) = 0

    E[X_1 X_2] = E[X_1] E[X_2]
  • Independent implies Uncorrelated
  • Uncorrelated does not necessarily imply Independent
    Only for linear functions of the form g(x) = a+bx, E[g(X_1)g(X_2) = E[g(X_1)]E[g(X_2)]
    Independence requires it to hold for all functions not just linear functions.
  • Independent ⊂ Uncorrelated
  • Example: Intra-day stock returns are dependent (not independent) and correlated (not uncorrelated, Cov ≠ 0).
    Inter-day Stock returns (from yesterday and today) are uncorrelated but not independent (still dependent).

Conditional Distribution

    F_{X_1|X_2}(x_1|x_2) = P(X_1 \leq x_1 | X_2 = x_2) = \left(\frac{\partial}{\partial x_2} F(x_1,x_2)\right)/f_{X_2}(x_2)

Conditional Density

    f_{X_1|X_2}(x_1|x_2) = \frac{\partial}{\partial x_1} F_{X_1|X_2}(x_1|x_2) = \frac{f(x_1,x_2)}{f_{X_2}(x_2)}
    \implies f_{X_1|X_2}(x_1|x_2) f_{X_2}(x_2) = f(x_1,x_2)
    f_{X_1}(x_1) = \int_{-\infty}^{\infty} f_{X_1|X_2}(x_1|x_2) f_{X_2}(x_2) dx_2

Conditional Expectation

  • E[g(X_1)|X_2] = \int_{-\infty}^{\infty} g(x_1) f_{X_1|X_2}(x_1|x_2) dx_1
  • E[g_1(X_1)g_2(X_2)|X_2] = g_2(X_2)E[g_1(X_1)|X_2]
    E[g(X_2)|X_2] = g(X_2)
  • E[E[g(X_1)|X_2]] = E[g(X_1)]

If X1 and X2 are independent,

    E[g(X_1)|X_2] = E[g(X_1)]

Conditional Variance

    \text{Var}[X_1|X_2] \equiv E[(X_1-E[X_1|X_2])^2|X_2]
    \text{Var}[X_1] = E[\text{Var}(X_1|X_2)] + \text{Var}(E[X_1|X_2])
    E[\text{Var}(X_1|X_2)] \leq \text{Var}[X_1]

Written by curious

August 9, 2010 at 9:57 am

Posted in statistics

STAT: Pareto Distribution

Written by curious

August 6, 2010 at 1:41 pm

Posted in statistics

STAT: Weibull Distribution

Weibull Distribution PDF

a=5, \gamma=2

Written by curious

August 6, 2010 at 1:32 pm

Posted in statistics