Quantitative Analysis

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Mea Variace Skewess Kurtosis Mea: X i = i Mode: Value that occurs most frequetly Media: Midpoit of data arraged i ascedig/ descedig

of squared deviatios from mea Sample variace i = = (Xi - X ( -) mea Populatio variace N = (X - ) i i µ σ = N Stadard deviatio = Variace

stock P= 00.0 σ of retur of stock Q=5.0 Cov (P,Q) =53. Curret Holdig $ m i P.

σ P = [w σ A + (-w) σ B +w(-w)cov(a,b)] w= 0.75 c = 00*(0.75) + 5*(0.5) +*0.5*0.75*53. σ P = 9.

3 Mea Variace Skewess Kurtosis Mea: X i = i Mode: Value that occurs most frequetly Media: Midpoit of data arraged i ascedig/ descedig order s Avg. of squared deviatios from mea Sample variace i = = (Xi - X ( -) mea Populatio variace N = (X - ) i i µ σ = N Stadard deviatio = Variace Var(ax+by)=a Var(x)+ b Var(y)+abCov(x,y) ) Positively: mea> media> mode Negatively: mea< media< mode Skewess of Normal = 0 σ of retur of stock P= 00.0 σ of retur of stock Q=5.0 Cov (P,Q) =53. Curret Holdig $ m i P. New Holdig: shiftig $ millio i Q ad keepig USD 3 millio i stock P. What %age of risk (σ), is reduced? σ P = [w σ A + (-w) σ B +w(-w)cov(a,b)] w= 0.75 c = 00*(0.75) + 5*(0.5) +*0.5*0.75*53. σ P = 9.5 old σ = 00 = 0 Reductio = 5% Leptokurtic: More peaked tha ormal (fat tails); kurtosis>3 Platykurtic: Flatter tha a ormal; kurtosis<3 Kurtosis of Normal = 3 Excess Kurtosis = Kurtosis - 3 If s of returs from fiacial istrumets are leptokurtotic. How does it compare with a ormal of the same mea ad variace? Leptokurtic refers to a with fatter tails tha the ormal, which implies greater kurtosis 3

Properties Coutig priciples Sum rule ad Bayes' Theorem Discrete Cotiuous P(A) = # of fav.

B)= P(A B)/P(B) P(A B)=P(A B)P(B) P(A B)=P(A)*P(B)If A,B idepedet No. of ways to select r out of objects: C r =!/[r!

Its expected aual default rate of 0%. Assume a costat quarterly default rate.

P(No Default Year) = P(No default i all Quarters) = (-PDQ)*(-PDQ)*(-PDQ3)*(-PDQ4) PDQ=PDQ=PDQ3=PDQ4=PDQ P(No Def

4% P( B) = P( A B) + P( A c B) c c P ( B) = P( B / A)* P( A) + P( B / A )* P( A ) P(A/B)* P(B) P(B/A) = c c P(A/B)*

the subsidiary defaults. Calculate of a subsidiary & paret both defaultig. Paret has a PD =.5% subsidiary has PD of.

P(x)= C x *p x *(-p) -x Biomial Radom Variable E(X)=*p Var(X)=*p*(-p)=*p*q A portfolio cosists of 7 ucorrelated

is eve over the year, Calculate prob. of exactly bods defaultig i first moth? -moth default rate =- (-0.593) / = 0.

4 Properties Coutig priciples Sum rule ad Bayes' Theorem Discrete Cotiuous P(A) = # of fav. Evets/ # of Total Evets 0 < P(A) <, P(A c )=-P(A) P(AUB)=P(A)+P(B)-P(A B) =P(A)+P(B) If A,B mutually exclusive P(A B)= P(A B)/P(B) P(A B)=P(A B)P(B) P(A B)=P(A)*P(B)If A,B idepedet No. of ways to select r out of objects: C r =!/[r!* (-r)!] No. of ways to arrage r objects i places: P r =!/(-r)! ABC was ic. o Ja, 004. Its expected aual default rate of 0%. Assume a costat quarterly default rate. What is the probability that ABC will ot have defaulted by April, 004? P(No Default Year) = P(No default i all Quarters) = (-PDQ)*(-PDQ)*(-PDQ3)*(-PDQ4) PDQ=PDQ=PDQ3=PDQ4=PDQ P(No Def Year) = (-PDQ) 4 P(No Def Quarter) = (0.9) 4 = 97.4% P( B) = P( A B) + P( A c B) c c P ( B) = P( B / A)* P( A) + P( B / A )* P( A ) P(A/B)* P(B) P(B/A) = c c P(A/B)* P(B) + P(A/B )*P(B ) The subsidiary will default if the paret defaults, but the paret will ot ecessarily default if the subsidiary defaults. Calculate of a subsidiary & paret both defaultig. Paret has a PD =.5% subsidiary has PD of.9% P(P S) = P(S/P)*P(P) = *0.5 = 0.5% Biomial Oly possible outcomes: failure or success. P(x)= C x *p x *(-p) -x Biomial Radom Variable E(X)=*p Var(X)=*p*(-p)=*p*q A portfolio cosists of 7 ucorrelated bods. The -year margial default prob. of each bod is 5.93%. If spread of default prob. is eve over the year, Calculate prob. of exactly bods defaultig i first moth? -moth default rate =- (-0.593) / = Ways to select bods out of 7 = 7 C = 7*6/ P(Exactly defaults) = (7*6/)*( ) *( ) 5 = 0.35% Poisso Fix the expectatio λ=p. P(x)=λ x e -λ /x! if x>=0 P(x)=0 otherwise AB The umber of false fire alarms i a suburb of Housto averages. per day. What is the (apprximate) probability that there would be 4 false alarms o day? P(X=x) = (λ x e -x )/x! X=., x = 4 P(.) = 0. 4

Discrete Cotiuous AB Cotiuous uiform Normal Distributio (ND) Outcome oly betwee [a, b] P(outside a & b) = 0 Cumulative desity fuctio (cdf)

X with desity fuctio f(x) = / (b - a) for a < x < b, ad 0 otherwise, is said to have a uiform over (a, b). Calculate its mea.

7% of Data -4-3 - - 0 3 4 If Z is a stadard ormal R.V. A evet X is defied to happe if either -< Z < or Z >.5. What is the prob.

5 The sum of areas show i two figures Area = -*(- N()) = -*(0.587) Area = 0.0668, Total Area = 0.

5 Discrete Cotiuous AB Cotiuous uiform Normal Distributio (ND) Outcome oly betwee [a, b] P(outside a & b) = 0 Cumulative desity fuctio (cdf) for Uiform : F(x)=0 For x <=a F(x)=(x-a)/(b-a) For a<x<b F(x)= For x >=b The R.V. X with desity fuctio f(x) = / (b - a) for a < x < b, ad 0 otherwise, is said to have a uiform over (a, b). Calculate its mea. a Sice the is uiform, the mea is the ceter of the, which is the average of a ad b = (a+b)/ 68% of Data 95% of Data 99.7% of Data If Z is a stadard ormal R.V. A evet X is defied to happe if either -< Z < or Z >.5. What is the prob. of evet X happeig if N () =0.843, N (0.5) = ad N (-.5) = , where N is the CDF of a stadard ormal variable The sum of areas show i two figures Area = -*(- N()) = -*(0.587) Area = , Total Area = Stadardized RV is ormalized mea = 0, σ = Z-score: # of σ a give observatio is from populatio mea. Z=(x-µ)/σ At a particular time, the market value of assets of the firm is $00 M ad the market value of debt is $80 M. The stadard deviatio of assets is $ 0 M. What is the distace to default? z = (A-K)/σ A = (00-80)/0 = 5

Cetral limit theorem SE (σ x ) of the sample mea is σ of the dist. of sample meas Kow pop.

s x = s/ () 5 observatio are take from a sample of kow variace.

You wish to coduct a two - tailed test of ull hypothesis that the mea is equal to 50.

6 Cetral limit theorem SE (σ x ) of the sample mea is σ of the dist. of sample meas Kow pop. Var. σ x = σ/ () Ukow pop. Var. s x = s/ () 5 observatio are take from a sample of kow variace. Sample mea =70 ad populatio σ = 60. You wish to coduct a two - tailed test of ull hypothesis that the mea is equal to 50. What is most appropriate test statistic? Stadard Error of mea (σ x ) = σ/ () = 60/ 5 = Degrees of freedom = 4 Use t- statistic = (x - μ)/ σ x = (70-50)/ =.67 As Sample Size icreases Distributio Becomes Almost Normal regardless of shape of populatio 6

Null hypothesis: H 0 Alterative : H a Oe tailed test Two tailed test Z & t test P- value Mea Test that the

rejectio of H 0 whe it is actually true Type II error:fail to reject H 0 whe it is actually false Real State of

error P (Type II error) = β Correct decisio Power = -β Cocluded if there is sigificat evidece to reject H 0 - -5

H a : µ>k $9,000 Critical value Test if the value is differet from K H 0 : µ=0 vs. H a : µ 0 α= 0.

05 Give H 0 true, of obtaiig value of test statistic at least as extreme as the oe that was actually observed H 0

7 Null hypothesis: H 0 Alterative : H a Oe tailed test Two tailed test Z & t test P- value Mea Test that the researcher wats to reject Iferece Based o Sample Data H 0 is True H 0 is False Actually tested Type I error: rejectio of H 0 whe it is actually true Type II error:fail to reject H 0 whe it is actually false Real State of Affairs H 0 is True Correct decisio Cofidece level = - α Type I error Sigificace level = α* H 0 is False Type II error P (Type II error) = β Correct decisio Power = -β Cocluded if there is sigificat evidece to reject H Test if the value is greater tha or less tha K H 0 :µ<=k vs. H a : µ>k $9,000 Critical value Test if the value is differet from K H 0 : µ=0 vs. H a : µ 0 α= Reject H μ c -μ =$,000 Do ot reject H 0 If <30 ad ukow σ, use t -Test Reject H 0 α= 0.05 Give H 0 true, of obtaiig value of test statistic at least as extreme as the oe that was actually observed H 0 : µ = µ vs H a : µ µ Use followig t-statistic for uequal variaces ( x x) ( µ µ ) t = ( s / ) + ( s / ) *Term α represets the maximum probability of committig a Type I error 7

00e (0.-.96*0.) < S < 00e (0.+.96*0.) 74.68 < S < 63.56 If stadard deviatio of a ormal populatio is kow to be 0 ad the mea is hypothesized to be 8. Suppose a sample size of 00 is cosidered.

8 Null hypothesis: H 0 Alterative : H a Oe tailed test Two tailed test Z & t test P- value Mea Test A stock has iitial price of $00. It price oe year from ow is give by S = 00 *exp(r), where the rate of retur r is ormally distributed with mea of 0. ad a stadard deviatio of 0.. What is the rage of S i a year with 95% cofidece? 00e (0.-.96*0.) < S < 00e (0.+.96*0.) < S < If stadard deviatio of a ormal populatio is kow to be 0 ad the mea is hypothesized to be 8. Suppose a sample size of 00 is cosidered. What is the rage of sample meas i which hypothesis ca be accepted at sigificace level of 0.05? s x = σ/ = 0/ 00 = z = (x-µ)/ σ x = (x-8)/ At 95% -.96<z<.96 ; So 6.04<x<9.96 Tests for a Sigle Populatio Variaces Chi-Square test H 0 : σ = c H a : σ c ( )s χ = σ Do ot reject H 0 Reject H χ 0 α α Upper tail test: H 0 : σ σ 0 H A : σ > σ 0 Tests for Variaces χ Tests for a two Populatio Variaces Do ot reject H 0 F test H 0 : σ σ = 0 H A : σ σ 0 s F = s H 0 : σ σ = 0 H A : σ σ 0 α/ F α/ Reject H 0 F 8

Simple Liear coefficiet Coefficiet of Determiatio(R ) Correlatio Coefficiet (CC) Residual Diagostic

X i = idepedet variable b 0 =itercept term; represets Y if X = 0 b = slope coefficiet; measures chage i Y

reject H 0 if t>+t critical or if t< -t critical %age of total var.

correlatio, - < CC <, if CC = 0, X & Y are ucorrelated r x,y = cov(x,y)/σ x σ y = R The error variable

9 Simple Liear coefficiet Coefficiet of Determiatio(R ) Correlatio Coefficiet (CC) Residual Diagostic Multiple LR model: Y i =b 0 +b X i +E i Y i = Depedet variable, estimated value of Y i, give value of X i X i = idepedet variable b 0 =itercept term; represets Y if X = 0 b = slope coefficiet; measures chage i Y for uit chage i X Appropriate Test structure: H 0 :b =0; H a :b 0 Test: t b =(b^-b )/s b^ Decisio Rule: reject H 0 if t>+t critical or if t< -t critical %age of total var. i Y explaied by X R =(SSR / SST) = -(SSE / SST) = explaied variatio/ total variatio Oly the liear correlatio, - < CC <, if CC = 0, X & Y are ucorrelated r x,y = cov(x,y)/σ x σ y = R The error variable must be ormally distributed, The error variable must have a costat variace The errors must be idepedet of each other Y = b + b X + b X b X i 0 i Adjusted R- square is used to test the goodess of fit R a i ki ( ) = R k k + ε i Coefficiets Stadard Error t-statistic Itercept X Variable X Variable X Variable

EWMA GARCH Implied Mote Carlo Advatages of simulatio modelig σ λσ = + ( λ) u Where, λ=persistece

= 0.95 to develop a forecast of the coditioal variace, which weight will be applied to the retur

For t=4, ad processig r 0 through the equatio three times produces a factor of (-0.95)*0.953 = 0.

variace= ω/ (-α-β) α+β+γ= α+β< for stability so that γ is ot -ve GARCH model is estimated as

000005 + µ t σ t The implied volatility of a optio cotract is the volatility implied by the market

estimate for ext day VL =.07%, Also, variace estimate for t+ =.000005 + 0.*(-%) + 0.85*(.88%) = 0.

78% Log Term I the GARCH model, % is the weight give to latest squared retur (reactive factor).

-0.85 = 3% is weight give to log-term average. Therefore, 3%*VL = 0.000005 i.e. VL = 0.

ad radomly selectig values from them, it recalculates the simulated model may times ad brigs out

10 EWMA GARCH Implied Mote Carlo Advatages of simulatio modelig σ λσ = + ( λ) u Where, λ=persistece factor/decay Factor - λ= Reactive factor Usig a daily RiskMetrics EWMA model with a decay factor λ = 0.95 to develop a forecast of the coditioal variace, which weight will be applied to the retur that is 4 days old? The EWMA RiskMetrics model is defied as h t = λ*h t - + (- λ)*r t -. For t=4, ad processig r 0 through the equatio three times produces a factor of (-0.95)*0.953 = for r 0 whe t = 4 σ ω αu βσ = + + ω =Weighted log ru variace= γvl VL=Log ru avg. variace= ω/ (-α-β) α+β+γ= α+β< for stability so that γ is ot -ve GARCH model is estimated as follows: σ = t µ t σ t The implied volatility of a optio cotract is the volatility implied by the market price of the optio based o a optio pricig model O a particular day 't'; actual retur was -% & the std. deviatio estimate was.8%. Calculate the volatility estimate for ext day (t+) ad log-term average volatility. estimate for ext day VL =.07%, Also, variace estimate for t+ = *(-%) *(.88%) = 0.037% (std. deviatio) estimate for t+ = sqrt(0.037%) =.78% Log Term I the GARCH model, % is the weight give to latest squared retur (reactive factor). 85% is the weight give to latest variace estimate (persistece factor). Therefore, = 3% is weight give to log-term average. Therefore, 3%*VL = i.e. VL = 0.07% Techique that coverts ucertaities i iput variables of a model ito probability s Combiig the s ad radomly selectig values from them, it recalculates the simulated model may times ad brigs out the probability of the output Techiques for Geeratig good radom umbers: Pseudoradom umber geerator Quasiradom umber geerators Stratified samplig Whe the iput radom variable follows some complex whe the output is a complex fuctio of iput variable simulatio modelig ca compoud probability whe there are multiple iput radom variables Correlatio betwee iput variables is take ito accout 0

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Quantitative Analysis

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