Conditional independence of distinct medical tests Let $V =$ {Person $A$ is affected by a specific virus} and $+ =$ {Person $A$ has tested positive for the virus}. Let also $T_1$ and $T_2$ be two distinct occurences of the same medical test (positive or negative). We are given that $P(V|+) = p,\ P(+|V) = q$ and $P(V) = v$.
Can we prove, using only the information specified above, that $T_1$ and $T_2$ are independent given $V$, namely
$$P(T_1 \cap T_2|V) = P(T_1|V)P(T_2|V)$$
For instance if $T_1$ and $T_2$ are both positive we would like to prove that $P(T_1 \cap T_2|V) = q^2$.
If not, under what conditions it can be reasonably assumed?
Edit:
To give some more context, my question arises from an exercise that was asking to find the probability that a person gets a positive first test and a negative second, given that he has a specific virus. The author solved the exercise assuming conditional independence of the two tests, so I wondered if there is a proof for that.
 A: You really need to know something about the biology of the disease and the inner workings of the test.
Following the comments on the other answer we work with a sensitivity (i.e. $P(+|V)$) of 90%.
Here are two, somewhat unrealistic scenarios:

*

*Suppose that 90% of people when infected with the virus produces some special anti-gen that the test picks up with 100% sensitivity, but 10% of the people have a genetic disorder that makes that they don't produce the antigen. Then from the manufacturers perspective the sensitivity of the test is 90%: giving the test to a random person from the set of all virus carriers yields 90% chance of a positive test result.
But the probability of getting 2 positive tests when testing the same infected person twice is still 90% because we are only testing if this person belongs to the 90% of the population that produces the anti-gen when infected.



*Suppose that 90% of tests are 100% sensitive to the virus, and 10% of tests are defective and are 0% sensitive. Now from a consumer perspective the sensitivity is 90%: if I have the virus and do a random test I have 90% chance of testing positive.
But the probability of testing positive twice when doing two tests of this type is suddenly only 81% because here we can assume conditional independence: buying a test is like picking a marble from a vase in one of those math problems.

Most realities will be somewhere in between these extremes. There is also dependence on how far you stick the test into your nose and a lot of other factors. The more you know, the better you can model the part that you don't know as the result of a random process. But what is the right model and how much independence it contains depends on what you do know about the underlying mechanisms.
A: "For instance if $T_1$ and $T_2$ are both positive we would like to prove that $P(T_1 \cap T_2|V) = q^2$"
In general,we certainly can't agree to  assertions of this type.
I would say we could only make that assertion if both the sensitivity and specificity of the two tests are identical or near identical.
[ Sensitivity $= \frac{True\;positive}{True\;positive + False\; negative}$,
Specificity $ = \frac{True\; negative}{True\;negative + False\; positive}\quad\quad$]
Such situations are quite unlikely, that is why so called "gold standards" exist for different tests

P.S.

*

*Please don't change your question after answer(s) have been received as it can invalidate  posted answer(s)


*$P(T_1\cap T_2 |V)$ is definitely not equal to $q^2$ as it implies that if the repeat test returns +, you are  less sure that you have the disease.


*It is not clear to me what you are driving at.
PPS:
It is best to take a simpler model where the test has an "accuracy", of , say, $90\,$% i.e. fraction of diseased people testing positive is $0.9$ and fraction of healthy people testing negative is also $0.9$.
Then, if two tests return positive,
it doesn't mean that P(diseased|positive twice) $= 0.9\cdot0.9$,
rather the probability is $1 -(1-0.9)^2$
In all the cross-correspondence, I had lost track of your  original query, viz if two (same) tests give values of $p_1$ and $p_2$ for P(+|diseased), would the value be $p_1\cdot p_2$ ? No, it would be the simple  average $\large\frac{p_1+p_2}{2}$
