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3. Finally, the probability of picking two tickets of each colour is

                                                                            2
                                                                       ( ) ( ) ( )       2
                                                                                   2
                                                                6!       1      1     1       10
                                    P(two of each colour) =          =                     =    .
                                                              2!2!2!     3      3     3       81
                     Thus the expected return to any patron was, in pence,
                                                    (          )
                                                       1     4           10
                                                100       +       + 40 ×     = 10.29.
                                                      243    81          81
                     A good time was had by all but the stallholder!


               The multivariate Gaussian distribution. A particularly interesting multivariate distribution is
               provided by the generalisation of the Gaussian distribution to multiple random variables X i , i = 1,
               2, . . . , n. If the expectation value of X i is E(X i ) = µ i then the general form of the PDF is given by
                                                           [                                ]
                                                              1  ∑ ∑
                                f(x 1 , x 2 , . . . , x n ) = N exp −   a ij (x i − µ i )(x j − µ j ) ,
                                                              2
                                                                 i   j
               where a ij = a ji and N is a normalization constant that we give below. If we write the column
                                                                T
                                        T
               vectors x = (x 1 x 2 · · · x n ) and µ = (µ 1 µ 2 · · · µ n ) , and denote the matrix with elements a ij by A
                                                                 T
                                                         1
               then f(x) = f(x 1 , x 2 , . . . , x n ) = N exp[− (x − µ) A(x − µ), where A is symmetric.
                                                         2
                   We can find that
                                                      2
                                                    ∂ M(0, 0, . . . , 0)
                                                                                  −1
                                        E(X i X j ) =                 = µ i µ j + (A ) ij ,
                                                          ∂t i ∂t j
               and thus, using (7.11), we obtain
                                                                                   −1
                                       Cov[X i , X j ] = E[(X i − µ i )(X j − µ j )] = (A ) ij .

               Hence A is equal to the inverse of the covariance matrix V of the X i , see (7.12). Thus, with the
               correct normalization, f(x) is given by

                                                    1               1
                                                                             T
                                    f(x) =                   exp[− (x − µ) V    −1 (x − µ)].              (7.16)
                                            (2π) n/2 (detV ) 1/2    2
                   In particular, we note that the Y i are independent Gaussian variables with mean zero and
               variance λ i . Thus, given a general set of n Gaussian variables x with means µ and covariance
               matrix V, one can always perform the above transformation to obtain a new set of variables y,
               which are linear combinations of the old ones and are distributed as independent Gaussians with
               zero mean and variances λ i .


                                 1.8. The Law of Large Numbers





                     Markov and Chebyshev’s inequalities


               Theorem 8.1. (Markov’s inequality.) If a random variable X accepts only non-negative
               values and has a finite mean, then for any positive number α the following inequality holds
                                                                   E(X)
                                                     P(X ≥ α) ≤          .                                 (8.1)
                                                                     α
                                                                                                              ⋆

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