Lupine Publishers| Journal of Biostatistics & Biometrics
Mini Review
In this brief description, we will use the numerator layout [1], and will tacitly assume that all products are conformable.
The derivative of the linear form π°tπ± with respect to the vector π±is given as
and since π°tπ± is a scalar, we are facing a particular case of the derivative of a scalar Ξ» with respect to a vector, e.g., ∂π±Ξ»=(∂π±1Ξ» ..... ∂π±nΞ») and it must also be ∂π±(π°tπ±)=∂π±(π±tπ°) . Moreover, it is easy to demonstrate that using the denominator layout, the derivative would have been ∂π±(π°tπ±)=π°
If both π° and π± vectors are function of a third vector π΅, we get
which, in the case π°=π±=π² reduces to
Dealing with a linear transform π°=aπ±, if A is π×π we have
and if π± is a function of a vector π²π± we get
From definition of bilinear form, we obtain, for π°tAπ± the derivative
while, for a quadratic form π°tAπ±(where A is π × π), we get
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