MIT6_041F10_assn10_sol

# MIT6_041F10_assn10_s - parenleftbigg parenrightbigg parenleftbigg parenrightbigg radicalBig parenleftBigg parenrightBigg summationdisplay

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Unformatted text preview: parenleftbigg parenrightbigg parenleftbigg parenrightbigg radicalBig parenleftBigg parenrightBigg summationdisplay summationdisplay summationdisplay parenleftbigg parenrightbigg Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Problem Set 10 Solutions 1. A financial parable. (a) The bank becomes insolvent if the asset’s gain R ≤ − 5 (i.e., it loses more than 5%). This probability is the CDF of R evaluated at − 5. Since R is normally distributed, we can convert this CDF to be in terms of a standard normal random variable by subtracting away the mean and dividing by the standard deviation, and then look up the value in a standard normal CDF table. E [ R ] = 7 , var( R ) = 10 2 = 100 , P ( R ≤ − 5) = P R − 7 ≤ − 5 − 7 = Φ( − 1 . 2) ≈ . 115 . 10 10 Thus, by investing in just this one asset, the bank has a 11.5% chance of becoming insolvent. (b) If we model the R i ’s as independent normal random variables, then their sum R = ( R 1 + + R 20 ) / 20 is also a normal random variable (see Example 4.11 on page 214 of the text). · · · Thus, we can calculate the mean and variance of this new R and proceed as in part (a). Note that since the random variables are assumed to be independent, the variance of their sum is just the sum of their individual variances. E [ R ] = ( E [ R 1 ] + + E [ R 20 ]) / 20 = 7 , · · · 1 20 100 var( R ) = (var( R 1 ) + + var( R 20 )) = · = 5 , 20 2 · · · 400 P ( R ≤ − 5) = P R √ − 5 7 ≤ − 5 √ − 5 7 = Φ( − 5 . 367) ≈ . 0000000439 = 4 . 39 · 10 − 8 . Thus, by diversifying and assuming that the 20 assets have independent gains, the bank has seemingly decreased its probability of becoming insolvent to a palatable value. (c) Now, if the gains R i are positively correlated, then we can no longer sum up the individual variances; we need to account for the covariance between pairs of random variables. The covariance is given by 1 cov( R i , R j ) = ρ ( R i , R j ) var( R i )var( R j ) = √ 10 2 10 2 = 50 . 2 · From page 220 in the text, we know that the variance in this case is 20 20 1 1 var( R ) = var R i = var( R i ) + cov( R i , R j ) 20 400 i =1 i =1 { ( i,j ) | i negationslash = j } 1 = (20 100 + 380 50) = 52 . 5 . 400 · · Since we assume that R = ( R 1 + + R 20 ) / 20 is still normal, we can again apply the same · · · steps as in parts (a) and (b): E [ R ] = ( E [ R 1 ] + + E [ R 20 ]) / 20 = 7 , · · · var( R ) = 52 . 5 , = P R − 7 = Φ( − 1 . 656) ≈ . 0488 . P ( R ≤ − 5) √ 52 . 5 ≤ − √ 5 52 − . 5 7 Page 1 of 7 parenleftBig parenrightBig parenleftBig parenrightBig Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Thus, by taking into account the positive correlation between the assets’ gains, we are no longer as comfortable with the probability of insolvency as we thought we were in part (b). longer as comfortable with the probability of insolvency as we thought we were in part (b)....
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## This note was uploaded on 01/11/2012 for the course EE 6.431 taught by Professor Prof.dimitribertsekas during the Fall '10 term at MIT.

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MIT6_041F10_assn10_s - parenleftbigg parenrightbigg parenleftbigg parenrightbigg radicalBig parenleftBigg parenrightBigg summationdisplay

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