intersection normal distributions and minimal decision error

Question

Assume controls $H_0 \sim \mathcal{N}(\mu_0, \sigma_0)$ and cases $H_1 \sim \mathcal{N}(\mu_1, \sigma_1)$, where $\mu_0 < \mu_1$ and $\sigma_0 = \sigma_1$. Let $c_1$ be the only intersection of the two distributions, i.e.,

$$f_H{_0}(c_1)−f_{H_1}(c_1)=0$$ Figure 1 shows an example. bi-normal curves.

Let $\alpha$ be the area of the $H_0$-distribution to the right of a decision threshold (False Positive errors) and $\beta$ be the area of the H1 distribution to the left of the decision threshold (False Negative errors).

Let c be the point where the minimum of $(\alpha + \beta)$ is reached $$\min(\alpha + \beta)$$ $$=\min[(1-P_{H_0}(H_0 < c)+P_{H_1}(H_1 < c)]$$ $$=\min[\int^c_{-\infty}(f_{H_1}(t))dt-\int^c_{-\infty}(f_{H_0}(t))dt+1]$$

Simulations in R showed that $c = c_1$ (at least, very near). How can you prove (or disprove) that $c = c_1$?

N.B. Schisterman, E. F., Perkins, N. J., Liu, A., & Bondell, H. (2005). Optimal cut-point and its corresponding Youden Index to discriminate individuals using pooled blood samples. Epidemiology, page 73–81, gave prove for a closely related problem, which I cannot reproduce (c.q., fail to understand).

Answer 1

In the case of a single threshold:

If c < c1, then $c \in (- \infty, c1)$

$$\int_{-{\infty}}^{c1} f_{H1}(t)dt-f_{H0}(t)dt= \int_{-{\infty}}^{c} f_{H1}(t)dt-f_{H0}(t)dt + \int_{c}^{c1} f_{H1}(t)dt-f_{H0}(t)dt$$ $$\int_{-{\infty}}^{c} f_{H1}(t)dt-f_{H0}(t)dt= \int_{-{\infty}}^{c1} f_{H1}(t)dt-f_{H0}(t)dt-\int_{c}^{c1} f_{H1}(t)dt-f_{H0}(t)dt$$ For values < c1 $f_H0 > f_H1 \Rightarrow -\int_c^{c1} f_{H1}(t)dt-f_{H0}(t)dt > 0$

Therefore, the minimum is reached when c = c1.

Likewise, if c > c1, then $c1 \in (- \infty, c)$ $$\int_{-\infty}^{c} f_{H1}(t)dt-f_{H0}(t)dt= \int_{-\infty}^{c1} f_{H1}(t)dt-f_{H0}(t)dt + \int_{c1}^{c} f_{H1}(t)dt-f_{H0}(t)dt$$

For values > c1 $f_H1 > f_H0 \Rightarrow \int_c^{c1} f_{H1}(t)dt-f_{H0}(t)dt > 0$

Therefore, $min(\int_{-\infty}^{c} f_{H1}(t)dt-f_{H0}(t)dt)$ is reached when c = c1.

Any comment is welcome.

Answer 2

In the case of two thresholds ($\mu_1 > \mu_0$ and $\sigma_0 \neq \sigma_1$) fig 2:

(In many cases, the second intersection is irrelevant, as its densities of both distributions are close to 0. In other cases (such as in fig 2), two intersections weaken the test, prevent a simple interpretation and would logically lead to the application of two thresholds. Nevertheless, in diagnostic instruments the existence of such a second intersection is often ignored, and a single threshold is applied. The probable reason for this is that in classical detection tools such as ROC curves, intersections are invisible.)

Let c1 and c2 respectively be the left and the right intersections (c2 > c1).

When $σ_1 < σ_0$ and for a threshold c, when c < c1 → $f_{H1} < f_{H0}$, when c1 > c > c2 → $f_{H1} > f_{H0}$, and when c > c2 → $f_{H1} < f_{H0}$. Furthermore, ($f_{H0}(c1) = f_{H1}(c1)) > (f_{H0}(c2) = f_{H1}(c2))$.

Let c be the minimum of $(α + β)$. If c < c1, the proof that $min(\int_{-\infty}^c (f_{H1}(t)-f_{H0}(t)) dt)$ will be reached when c = c1, the proof is equal to that given for the single intersection case. Idem, if c1 > c > c2, the proof is equal to the case of a single intersection and c1 > c.

If c > c2 → c1 in (-∞, c2) and c2 in (-∞, c): $$\int_{-\infty}^c (f_{H1}(t)-f_{H0}(t)) dt = \int_{-\infty}^{c1} (f_{H1}(t)-f_{H0}(t)) dt+\int_{c1}^{c2} (f_{H1}(t)-f_{H0}(t)) dt+\int_{c2}^c (f_{H1}(t)-f_{H0}(t)) dt$$ As $\int_{c1}^{c2} (f_{H1}(t)-f_{H0}(t)) dt > 0$, and $\int_{c2}^c (f_{H1}(t)-f_{H0}(t)) dt < 0$, but $\int_{c1}^{c2} (f_{H1}(t)-f_{H0}(t)) dt > \int_{c2}^c (f_{H1}(t)-f_{H0}(t)) dt$, therefore $\int_{c1}^{c2} (f_{H1}(t)-f_{H0}(t)) dt + \int_{c2}^c (f_{H1}(t)-f_{H0}(t)) dt > 0$, and consequently the minimum will be reached when c = c1.

When $σ_1 > σ_0$ and for a threshold c, when c < c1 → $f_{H1} > f_{H0}$, when c1 > c > c2 → $f_{H1} < f_{H0}$, and when c > c2 → $f_{H1} > f_{H0}$. Furthermore, ($f_{H0}(c1) = f_{H1}(c1)) < (f_{H0}(c2) = f_{H1}(c2))$.The proof is along similar lines with some reversals, and the minimum is reached for intersection c2 instead of c1.

In both cases, the other intersection is where the sum of the left tail of H0 and the right tail of H1 is minimized, the opposite of what we want when $µ_1 > µ_0$.

This proof is dependent on visual inspection, especially the dependency of the relative pattern on the sigma parameter. The next question is whether the same also holds for other kinds of distributions. When the same relative pattern of the two distributions occur, this seems to be true. Schisterman et al. seem to assume that it holds for Gamma distributions, which is probably true for the distributions they consider, and which have forms that do not differ too much from normal distributions. I wonder whether it also holds for distributions that differ more considerably. Clearly, the relative pattern of two non-normal distributions, would be dependent on the parameters of the non-normal distributions.

Any suggestion, comment or help is welcome.