ln() and product confuse meI have this function: p ( x n ; μ ) = 1 ( 2 π ) l 2 | Σ | 1 2 exp ⁡ ( − 1 2 ( x n − μ ) T Σ − 1 ( x n − μ ) ) And I have to compute this: L ( μ ) = ln ⁡ ( ∏ n = 1 N p ( x n ; μ ) ) which should be equal to: − N 2 ln ⁡ ( ( 2 π ) l | Σ | ) − 1 2 ∑ n = 1 N ( x n − μ ) T Σ − 1 ( x n − μ )Note: Σ is a covariance matrix.From my attempt I have: 1 ( 2 π ) l 2 | Σ | 1 2 = ln ⁡ 1 − ln ⁡ ( ( 2 π ) l / 2 | Σ | 1 / 2 ) = − ln ⁡ ( ( 2 π ) l / 2 | Σ | 1 / 2 ) = − L 2 ln ⁡ 2 π − 1 2 ln ⁡ | Σ | and: ln ⁡ ( exp ⁡ ( − 1 2 ( x n − μ ) T Σ − 1 ( x n − μ ) ) ) = − 1 2 ( x n − μ ) T Σ − 1 ( x n − μ )but when I put everything together with the product, it doesn't work out, squares appear and it doesn't feel right...Can you help me please understand?

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ln() and product confuse meI have this function:  p  (      x    n    ;  μ  )  =      1          (      2      π              )                              l            2                                      |            Σ                        |                                      1            2                                exp  ⁡      (    −          1      2        (          x      n        −    μ          )      T              Σ              −        1              (          x      n        −    μ    )    )  And I have to compute this:   L  (  μ  )  =  ln  ⁡      (          ∏              n        =        1                    N              p    (          x      n        ;    μ    )    )   which should be equal to:  −      N    2      ln  ⁡  (     (  2  π      )    l        |    Σ      |       )     −         1    2           ∑          n      =      1              N        (      x    n    −  μ      )    T        Σ          −      1        (      x    n    −  μ  )Note:   Σ is a covariance matrix.From my attempt I have:      1          (      2      π              )                              l            2                                      |            Σ                        |                                      1            2                                =  ln  ⁡  1  −  ln  ⁡  (  (  2  π      )          l              /            2            |    Σ            |              1              /            2        )  =  −  ln  ⁡  (  (  2  π      )          l              /            2            |    Σ            |              1              /            2        )  =  −      L    2    ln  ⁡  2  π  −      1    2    ln  ⁡      |    Σ      |  and:  ln  ⁡  (  exp  ⁡  (  −      1    2    (      x    n    −  μ      )    T        Σ          −      1        (      x    n    −  μ  )  )  )  =  −      1    2    (      x    n    −  μ      )    T        Σ          −      1        (      x    n    −  μ  )but when I put everything together with the product, it doesn&#039;t work out, squares appear and it doesn&#039;t feel right...Can you help me please understand?

zepplinkid7yk · Accepted Answer

We are given that  p  (      x    n    ;  μ  )  :=      1          (      2      π              )                              l            2                                      |            Σ                        |                                      1            2                                exp  ⁡      (    −          1      2        (          x      n        −    μ          )      T              Σ              −        1              (          x      n        −    μ    )    )  Since   p is real valued we should be able to apply ordinary log rules within reason. These are:1.  ln  ⁡  (  x  y  )  =  ln  ⁡  (  x  )  +  ln  ⁡  (  y  ) which generalises to   ln  ⁡      (          ∏      i              x      i        )    =      ∑    i    ln  ⁡  (      x    i    ) for finite sums and products.2.  ln  ⁡  (      a    b    )  =  b  ln  ⁡  (  a  )3.  ln  ⁡  (  exp  ⁡  (  y  )  )  =  yNow we want to find:  L  (  μ  )  =  ln  ⁡      (          ∏              n        =        1            N        p    (          x      n        ;    μ    )    )    =      ∑          n      =      1        N    ln  ⁡  (  p  (      x    n    ;  μ  )  )    using point   1.Now we find   ln  ⁡  (  p  (      x    n    ;  μ  )  )  :                    ln        ⁡        (        p        (                  x          n                ;        μ        )        )                            =        ln        ⁡                  (          (          2          π                      )                          −              l                              /                            2                                            |                    Σ                                    |                                      −              1                              /                            2                                exp          ⁡                      (            −                          1              2                        (                          x              n                        −            μ                          )              T                                      Σ                              −                1                                      (                          x              n                        −            μ            )            )                    )                                                  =        ln        ⁡                  (                                    [              (              2              π                              )                l                                            |                            Σ                              |                            ]                                      −              1                              /                            2                                )                +        ln        ⁡                  (          exp          ⁡                      (            −                          1              2                        (                          x              n                        −            μ                          )              T                                      Σ                              −                1                                      (                          x              n                        −            μ            )            )                    )                                                  =        −                  1          2                ln        ⁡                  (          (          2          π                      )            l                                |                    Σ                      |                    )                +                  (          −                      1            2                    (                      x            n                    −          μ                      )            T                                Σ                          −              1                                (                      x            n                    −          μ          )          )                                                  =        −                  1          2                ln        ⁡                  (          (          2          π                      )            l                                |                    Σ                      |                    )                −                  1          2                (                  x          n                −        μ                  )          T                          Σ                      −            1                          (                  x          n                −        μ        )            We used points 2 and 3 in going from line 2 to 3 above. Now to do the sum, note the first term doesn&#039;t depend on n                              ∑                      n            =            1                    N                ln        ⁡        (        p        (                  x          n                ;        μ        )        )                            =                  ∑                      n            =            1                    N                          [          −                      1            2                    ln          ⁡                      (            (            2            π                          )              l                                      |                        Σ                          |                        )                    −                      1            2                    (                      x            n                    −          μ                      )            T                                Σ                          −              1                                (                      x            n                    −          μ          )          ]                                                  =        −                  N          2                ln        ⁡                  (          (          2          π                      )            l                                |                    Σ                      |                    )                −                  1          2                          ∑                      n            =            1                    N                (                  x          n                −        μ                  )          T                          Σ                      −            1                          (                  x          n                −        μ        )            Which is what you want.

ln() and product confuse me I have this function: p ( x n </msub> ;

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