Finding a partial differential to binomial distribution

[axiomlu]

View attachment 9326

 

[HallsofIvy]

How many times are you going to post the same question? Well, maybe it is not the same! Previously, you said "binormial", which I had first interpreted as "binomial", which you have here, but apparently you meant "bi-normal", a normal distribution in two variables. I would expect "N" to be a normal distribution which has a single variable with the mean and standard deviation as

parameters. I don't know what you mean by "N(a, b, c, d, \rho)", with 5 variables. Also, since N reduces to a function in the single variable, you should get \frac{dN}{dx}, not \frac{\partial N}{\partial x} but that is a matter of notation, not substance.

In any case, by the "chain rule", \frac{dN}{dx}= \frac{\partial N}{\partial a}\frac{da}{dx}+ \frac{\partial N}{\partial b}\frac{db}{dx}+ \frac{\partial N}{\partial c}\frac{dc}{dx}+ \frac{\partial N}{\partial d}\frac{dd}{dx}+ \frac{\partial N}{\partial \rho}\frac{d\rho}{dx}.

Since a= 0.5x+ 3, da/dx= 0.5, b= -2x, db/dx= -2, c= x^2, dc/dx= 2x, d= x+ 0.2, dd/dx= 1, \rho= 0.4x- 0.2, d\rho/dx= 0.4 so
\frac{dN}{dx}= 0.5\frac{\partial N}{\partial a}- 2\frac{\partial N}{\partial b}+ 2x\frac{\partial N}{\partial c}+ \frac{\partial N}{\partial d}+ 0.4\frac{\partial N}{\partial \rho}.
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[I like Serena]

Hi axiomlu,

What does $N$ represent?
Normally it would represent a probability distribution, and more specifically the Normal Distribution.
However, that is not a function that we can take a derivative of.
And it is already clear that it is not the normal distribution, since that one has a single mean and a single variance.

Is it perhaps supposed to represent the probability density function of a distribution?
Or is it supposed to be something abstract. If so, HallsofIvy's response gives the best we can do.