(Why) Does positive predictive value depend on sensitivity? Assume there are two possible categories, coded as $0$ and $1$.
Define positive predictive value (PPV) to be the probability that an observation really is in category $1$ when it is predicted as being in category $1$. This is distinct from sensitivity or recall, which is the probability of predicting an observation to be in category $1$ when it is in category $1$. That is, if $y$ is the true observation and $\hat y$ is the prediction:
$$
P\left(
y = 1\vert \hat y=1
\right)\\
\text{Sensitivity} = P\left(
\hat y = 1\vert y=1
\right)
$$
To me, it seems like PPV should not care about the test sensitivity. PPV conditions on predicting category $1$. Therefore, no matter how remote the possibility is of predicting category $1$, PPV only takes over once such a prediction has been made.
At the same time, it seems like PPV and sensitivity should have some inverse relationship. If the test is skeptical and unlikely to predict category $1$, then a prediction of category $1$ should be regarded as quite remarkable and worth taking seriously. At the same time, if the test will call pretty much anything category $1$ (but doesn't miss cases, so high sensitivity), then who cares when it flags an observation as being in category $1?$ It does that all the time. That is not a special event, and there is not a particularly high probability that the observation really is in category $1$ just because the predictor makes such a prediction.
Applying Bayes' theorem to the PPV sticks sensitivity in the denominator, but I am wondering if my calculation can be taken further to reveal additional insight (particularly about specificity, which I take to be a measure of skepticism) or even get sensitivity to drop out entirely (which I doubt).
$$
\text{PPV} = P\left(
y = 1\vert \hat y=1
\right)=\dfrac{
P\left(
\hat y = 1\vert y=1
\right)P(y=1)
}{
P(\hat y = 1)
}$$$$=\dfrac{
P\left(
\hat y = 1\vert y=1
\right)P(y=1)
}{
P(\hat y = 1\vert y = 1) + P(\hat y = 1\vert y = 0)
}
$$

There is an error in your formula.   The denominator should be
$$P(\hat{y}=1|y=1)P(y=1) + P(\hat{y}=1|y=0)P(y=0)$$
What you have in the denominator is Sensitivity + (1 - Specificity), which is incorrect.
PPV can be problematic because it depends on the pre-test probability $P(y=1)$ as well as the sensitivity and specificity of the test.  Perfectly Specific tests when Positive rule In "disease" (i.e., $y=1$).  There's a mnemonic for this: SpPIn.  It goes along with SnNOut, which stands for perfectly Sensitive tests when Negative rule Out "disease".
Sensitivity and specificity tend to be local, stable characteristics of a test. They are less dependent on other factors besides $y=1$ vs. $y=0$, which is why they (and not positive and negative predictive value) are the “atomic” relationships in diagnostic reasoning.
For example, assume that you sample cases with $y=1$ separately from an equal number of controls with $y = 0$.  You can estimate sensitivity and specificity, then use these estimates along with a different $P(y=1)$ (other than 0.5) to get PPV and NPV.  If you try to estimate PPV from that same case-control design, you are assuming that wherever you apply the test, $P(y=1) = P(y=0) = 0.5$.
It is sometimes easier to think in terms of the odds form of Bayes's Rule:
\begin{align*}
P(y=1|\hat{y}=1)P(\hat{y}=1) &= P(\hat{y}=1|y=1)P(y=1) \\
P(y=0|\hat{y}=1)P(\hat{y}=1) &= P(\hat{y}=1|y=0)P(y=0) \\
\frac{P(y=1|\hat{y}=1)}{P(y=0|\hat{y}=1)} &= \frac{P(\hat{y}=1|y=1)}{P(\hat{y}=1|y=0)}\frac{P(y=1)}{P(y=0} \\
Odds(y |\hat{y}=1) &= LR(\hat{y}=1)Odds(y)\\
\end{align*}
Pre-test odds = $Odds(y) = \frac{P(y=1)}{P(y=0)}$
Post-test odds = $Odds(y|\hat{y}=1) = \frac{P(y=1|\hat{y}=1)}{P(y=0 | \hat{y}=1)}$
Likelihood ratio for $\hat{y} = r$ is $\frac{P(\hat{y}=r|y=1)}{P(\hat{y}=r|y=0}$
$$\text{Post-test Odds} = \text{Likelihood ratio}(r) \times \text{Pre-test Odds}$$
What we think now = New Information combined with What we thought before.
