Proving an equivalence relation by using numerical methods such as Gaussian quadrature

by Lang Wu   Last Updated April 07, 2017 07:19 AM

The following is my problem description.

Model one:
$$S_1(t)=\int \exp\left\{-\int_0^t h_0(t) \exp(\phi+\theta t) \, dt\right\} f(\theta) d\theta$$ Model two: $$S_2(t)=1-\int P(r(t)\ge Y\mid Y=y)p(Y=y) \, dy=1-\int F(t\mid y)f_Y(y) \, dy$$ Now I want to prove that model one is equivalent to model two, or to say, $S_1(t)=S_2(t).$
The idea is as follow.
Proposition: For any given model one, that is to say, $h_0(t)$ and $f(θ)$ are known in model one, we can always find an equivalent model two by manipulating its parameters such as the distribution of $Y$ $f_Y(y)$. On this occasion, we can assume that model two is a special case of model one.
To be specific, after getting $S_1(t)$ , by utilizing the equation $S_1(t)=S_2(t)$, we can obtain the distribution of $Y$ in model two on the condition that $F(t\mid y)$ is known. Here any restrictive parametric distribution form for $Y$ is not pre-assumed.


  1. I have ever tried the numerical integration method to obtain $f_Y(y)$ but failed. Due to the complexity of two models, I don’t know how to simplify the integration and thereby get the distribution of $Y$.

  2. Actually, I even needn’t figure out the specific expression of $f_Y(y).$ What I need is the method to prove the existence of $f_Y(y)$. Are there other ideas except for computing $f_Y(y)$? On the other hand, if I want to negate the proposition above, or to say, to demonstrate that $f_Y(y)$ is non-existent for the equivalence relation, what should I do?

I truly hope someone can provide me some suggestions or ideas. Thank you, all viewers!

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