In a common base configuration with a pnp and VBC≈0, why does an output current IC flow when VEB < 0

Suppose a common base configuration of a standard pnp bipolar-transistor with the base grounded. The idea is to use the transistor as an ideal diode (as by the Shockley equation):
$$
I_C = I_0 left(e^{e V_{EB}/kT}-1 right)
$$
Now, when I set $V_{EB}<0$, the equation tells us, that we will get a leakage current.

I don’t understand the physical reason behind that. Thus far I understand, that:
When operating in normal mode $(V_{EB}>0)$ holes are injected from the emitter into the base-collector depletion-region, where they are accelerated by the built-in potential and cause a collector current $I_C$. Now let’s say we apply a negative voltage to the emitter. Thus the built-in potential of the emitter-base diode increases and the diffusion current decreases. The current caused by thermal generation of electron-hole pairs however, will remain relatively constant and a reverse leakage current will flow. But how does this current affect the collector current? Charge carriers from thermal generation are majority carriers where they end up and should therefore not influence the other junction. Where is my mistake?

Other things I like to mention: I know, that in reality it is not possible to measure the collector current without also causing a potential difference between the collector and the base. I also know, that there is a base current and not all emitter current is flowing into the collector.