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Probability and Statistics for Finance (Frank J. Fabozzi Series) ( PDFDrive )
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Probability and Statistics

Discrete Case A formal definition of independence of any two discrete random variables Xi and Xj is given by (14.15) for all values a,b R. That is, we have to check that equation (14.15) is satisfied for any Xi value a in combination with any Xj value b. In case we should find a pair (a,b) that violates equation (14.15), then Xi and Xj are dependent. Note that here independence is only defined for two random variables. If we want to validate the independence of more than two random variables, we have to prove the fulfillment of equation (14.15) for any pair, as well as of the analog extension of equation (14.15) for any triple of these random variables, any set of four, and so on. That is, we have to validate
id: 91d80f4f3bcf57da06e07c29d2562aef - page: 467
Recall that in the example with the discrete asset random vector A = (A1,A2) that a portfolio value of $2,100,000 could not be achieved. This was the case even though the marginal distributions permitted the isolated values A1 = $110 and A2 = $120 such that $2,100,000 might be obtained or even exceeded. However, these combinations of A1 and A2 to realize $2,100,000, such as ($110,$120), for example, were assigned zero probability by the joint distribution such that they never happen. Let us validate the condition (14.15) for our two assets taking the value pair ($110,$120).157 The probability of this event is P($110,$120) = 0 according to the joint probability table given earlier in this chapter. The respective marginal probabilities are P(A1 = $110) = 0.45 and P(A2 = $120) = 0.5. Thus, we have P($110,$120) = 0 0.45 0.5 = P(A1 = $110) P(A2 = $120), and, consequently the asset values are dependent.
id: 6337f2f4a99bab9a83ef84c3f9f2f987 - page: 467
Continuous Case For the analysis of dependence of two continuous random variables Xi and Xj, we perform a similar task as we have done in equation (14.15) in the discrete case. However, since the random variables are continuous, the probability of single values is always zero. So, we have to use the joint and marginal density functions. In the continuous case, independence is defined as (14.16) is the joint density function of Xi and Xj.158 Again, if we find a pair where fXi, Xj (a,b) such that condition (14.16) is not met, we know that Xi and Xj, are dependent. We can extend the condition (14.16) to a k-dimensional generalization that, when X1, X2, , Xk are independent, their joint density function has the following appearance (14.17)
id: e00645b82fd50852f038744538e0aaf5 - page: 469
So if we can write the joint density function as the product of the marginal density functions, as done on the right side of equation (14.17), we conclude that the k random variables are independent. At this point, lets introduce the concept of convolution. Technically, the convolution of two functions f and g is the integral for any real number z. That is, it is the product of two functions f and g integrated over all real numbers such that for each value x, f is evaluated at x and, simultaneously, g is evaluated at the difference z - x. Now, lets think of f and g as the density functions of some independent continuous random variables X and Y, for example. Then, the convolution integral h yields the density of the
id: 2279d6364ae9d8d4fe05cfb9c1e68df6 - page: 469
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