I Likelihood of the maximum of a parabola

[dIndy]

I have a quadratic regression model $y = ax^2 + bx + c + \text{noise}$. I also have a prior distribution $p(a,b,c) = p(a)p(b)p(c)$. What I need to calculate is the likelihood of the data given solely the extremum of the parabola (in my case a maximum) $x_{max} = M = -\frac{b}{2a}$. What I tried so far is:

$$p(y|M) = \int p(y|M,b,c)p(b,c|M)\,dbdc$$

I would like to rewrite this as a function of $p(y|a,b,c)$ and $p(b,c|a) = p(b,c)$, substituting $a$ for $M$. However, I'm not sure how to perform a change of variables for conditional variables.

What I've also tried is using Bayes' theorem to rewrite the likelihood:
$$p(y|M) = \frac{p(y)p(M|y)}{p(M)} = \frac{p(y)\int p(M,b,c|y)\,dbdc}{\int p(M,b,c)\,dbdc} $$
Then performing the substitution for $M$:
$$ \frac{p(y)\int p(a,b,c|y)\det(J)\,dbdc}{\int p(a,b,c)\det(J)\,dbdc}$$
Using Bayes' theorem again:
$$ \frac{\int p(y|a,b,c)p(a,b,c)\det(J)\,dbdc}{\int p(a,b,c)\det(J)\,dbdc} = \frac{\int p(y|a,b,c)p(b,c)\det(J)\,dbdc}{\int p(b,c)\det(J)\,dbdc}$$

Can this be simplified further?

I've also got a far more challenging and general problem:

Given a non-linear regression model $y_i = f(\theta,x_i) + \text{noise}$, with $\theta$ the vector of unknown parameters and $x_i$ the vector of dependent variables. I want to calculate the likelihood of the global maximum of $f$. The problem here is that there is no closed form expression for $x_{max}$.

 

[tnich]

I have a quadratic regression model $y = ax^2 + bx + c + \text{noise}$. I also have a prior distribution $p(a,b,c) = p(a)p(b)p(c)$. What I need to calculate is the likelihood of the data given solely the extremum of the parabola (in my case a maximum) $x_{max} = M = -\frac{b}{2a}$. What I tried so far is:

$$p(y|M) = \int p(y|M,b,c)p(b,c|M)\,dbdc$$

I would like to rewrite this as a function of $p(y|a,b,c)$ and $p(b,c|a) = p(b,c)$, substituting $a$ for $M$. However, I'm not sure how to perform a change of variables for conditional variables.

What I've also tried is using Bayes' theorem to rewrite the likelihood:
$$p(y|M) = \frac{p(y)p(M|y)}{p(M)} = \frac{p(y)\int p(M,b,c|y)\,dbdc}{\int p(M,b,c)\,dbdc} $$
Then performing the substitution for $M$:
$$ \frac{p(y)\int p(a,b,c|y)\det(J)\,dbdc}{\int p(a,b,c)\det(J)\,dbdc}$$
Using Bayes' theorem again:
$$ \frac{\int p(y|a,b,c)p(a,b,c)\det(J)\,dbdc}{\int p(a,b,c)\det(J)\,dbdc} = \frac{\int p(y|a,b,c)p(b,c)\det(J)\,dbdc}{\int p(b,c)\det(J)\,dbdc}$$

Can this be simplified further?

I've also got a far more challenging and general problem:

Given a non-linear regression model $y_i = f(\theta,x_i) + \text{noise}$, with $\theta$ the vector of unknown parameters and $x_i$ the vector of dependent variables. I want to calculate the likelihood of the global maximum of $f$. The problem here is that there is no closed form expression for $x_{max}$.

It seems to me that you need a pdf $f_{\frac{-b}{2a}}(M)$ for $M$. Given pdfs for a for b, I think you could derive the pdf for $M$ as a convolution. Something like $p(M = \frac{-b}{2a}) = p(2aM + b = 0) = \int_{-\infty}^{\infty} p(a=\frac x {2M}) p(b = t-x)dx|_{t=0}$.

 

[Stephen Tashi]

. What I tried so far is:

$$p(y|M) = \int p(y|M,b,c)p(b,c|M)\,dbdc$$

It will be confusing if you use the notatiion $p(...)$ to denote both "probability of" and also a probability density function evaluated somewhere.

For example, using the notation $p_X()$ to denote the probability density function of the random variable $X$ we can write
$p_M(m) = \int p_A(a) p_B( -2am)\ da$. I don't know how you would write that using your notation.

Can we assume your priors assign zero probability to the case a=0?