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Integrating of an exponential of a matrix product

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I try to solve this integral with with parameter x as a member of this scale:(-∞ , +∞)
    I=∫∏dx exp(-0.5XAX + XB)=∫∏dx exp( Ʃ-0.5xa[j]x[j] +Ʃ xb )
    In which a[j] and b are components of telated matrix and vector and the first sum is on i and j ranges from 1 to N .Also X and B are two vector with N component and A is a N*N matrix,so the integral is over all x (which denote components of X).


    2. Relevant equations
    Gaussian integrals in the same scale obey this equation:
    ∫dx exp( -ax^2 ) =√(π/a)


    3. The attempt at a solution

    Using a change in variables like X=Y + CB with CA=Ac=1 should be appropriate:

    -0.5(XAX) + XB=(-0.5)(Y+CB)A(Y+CB)+(Y+CB)B=(-0.5)(YAY+YACB+CBAY+(CB)^2)+YB+CB^2
    = (-0.5)(YAY+YACB+CBAY+(CB)^2)+YACB+CB^2
    =(-0.5)(YAY+CBAY+(CB)^2-YACB + 2CBB)

    The problem is there is no YY=Ʃyy to help me change this problem to a Gaussian problem.Another problem is with YACB and CBAY that arenot compatible with this equation:
    (A+B)^2=A^2+B^2+AB+BA .

    Thank you for noticing.
     
  2. jcsd
  3. Mar 29, 2012 #2

    Ray Vickson

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    Your exponent is -(1/2)Q, where Q = <x,Ax> - 2<b,x> (writing the inner product sum u[j]*v[j] as <u,v>). We need to assume A is *symmetric*; if not, we can replace it by a new A that IS symmetric, and work with that. Writing x = y + c (c a constant vector) we have Q = <y+c,A(y+c)> -2<b,y+c> = <y,Ay> + <y,Ac> + <c,Ay> + <c,Ac> - 2<b,y> = <y,Ay> + <c,Ac> +<y,2Ac-2b>. If we take Ac=b we eliminate linear terms and just have Q = <y,Ay> + <c,Ac>.

    Now, we must assume A is positive-definite; otherwise, the integral will not be convergent. It follows that A is invertible, so c = Inverse(A)b can be found. Furthermore, we can perform a Cholesky factorization of A, to write A = U^T U, where U^T = transpose of U and U is upper-triangular with all u[i,i] > 0. Thus, if U[1] = u[1,1]y[1] + u[1,2]y[2] + ... + u[1,n]y[n], U[2] = u[2,2]y[2] + u[2,3]y[3] + ... + u[2,n]y[n],..., U[n] = u[n,n]y[n] we have <y,Ay> = U[1]^2 + U[2]^2 + ... + U[n]^2. We can change the variables of integration from y[1], y[2], ..., y[n] to U[1], U[2],...,U[n], to get:
    [tex] \int_{R^n} \exp{\left[-\frac{1}{2}<x,Ax> + <b,x>\right]} \, dx_1 \cdots dx_n
    = C\, \exp{\left(-\frac{1}{2}<c,Ac>\right)} \int_{R^n} \exp{\left[-\frac{1}{2} (U_1^2 + \cdots + U_n^2)\right]} \, dU_1 \cdots dU_n, [/tex]
    where C is the Jacobian you get by changing variables from y to U.

    RGV
     
    Last edited: Mar 29, 2012
  4. Apr 1, 2012 #3
    Please explain some points:

    1)Using X=Y+C,Q will Change this way:
    <X,AX>-2<B,X>=<Y+C,A(Y+C)>-2<B,Y+C>
    =<Y,AY>+<Y,AC>+<C,AY>+<C,AC>-2<B,Y>-2<B,C>
    I noticed that with the last sentence the final form for exponent -0.5Qwill produce
    <C,AC>/2 instead of -<C,AC>/2.

    2)It seems to me that :
    Q= <y,Ay> + <y,Ac> + <c,Ay> + <c,Ac> - 2<b,y> -2<b,c>
    = <y,Ay> + <c,Ac> +<c,Ay> + <y,Ac> - 2<b,y> -2<b,c>
    = <y,Ay> + <c,Ac> +<y,2Ac-2b> -2<b,c> - <y,Ac>
    By taking Ac=b we obtain Q= <y,Ay> + <c,Ac> - <y,Ac> -2<b,c>

    Please explain how you wrote Q = <y,Ay> + <c,Ac>

    FBR
     
  5. Apr 1, 2012 #4

    Ray Vickson

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    You are supposed to know that for real symmetric A we have [itex]<c,Ay> = <Ac,y> = <y,Ac> = <Ay,c>. [/itex] However, you are right in noting that I incorrectly dropped the term 2<b,c>, so we should have [itex] Q = <y,Ay> + <c,Ac> +2<y,Ac-b> - 2<b,c> = <y,Ay> + <c,Ac> - 2<b,c>[/itex] if we take [itex] Ac=b[/itex]. In fact, this gives [itex] <c,Ac> - 2<b,c> = -<b,c>, [/itex] so
    [tex] \int_{R^n} \exp(-Q/2) \, dx_1 \cdots dx_n = C\, \exp(-<b,c>) \int_{R^n} \exp(-(U_1^2 + \cdots + U_n^2)/2) \, dU_1 \cdots dU_n. [/tex]

    RGV
     
    Last edited: Apr 2, 2012
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