= Solution
Let $p_0=\overline\Phi(2)$ and $p_1=\overline\Phi(2/\sqrt{1+V})$, where $\overline\Phi$ is the standard <normal distribution> upper-tail probability. The <mixture model> gives
$$
\Pr(Y_i>2\mid q,V)=(1-q)p_0+qp_1.
$$
For $V>0$, <threshold inference for a mixture proportion> therefore gives
$$
\boxed{\widehat q=\frac{0.15-\overline\Phi(2)}
{\overline\Phi(2/\sqrt{1+V})-\overline\Phi(2)}.}
$$
This is a method-of-moments estimate when it lies in $[0,1]$. More formally, if $K$ of the $N$ independent genes exceed the threshold, then $K\sim\operatorname{Binomial}(N,(1-q)p_0+qp_1)$, and its constrained <maximum-likelihood estimate> clips the displayed value to $[0,1]$. No numerical estimate is determined without the given value of $V$.
Here $p_0\approx0.02275$. Matching $15\%$ exactly requires $p_1\geq0.15$, equivalently $V\geq[2/\Phi^{-1}(0.85)]^2-1\approx2.724$. For smaller positive $V$, the maximum is at $q=1$, and a large-sample exceedance fraction this high suggests a mismatch with the prescribed component variances. Sampling fluctuation is still possible; an observed fraction need not equal its expectation. If $V=0$, the denominator vanishes and this summary gives no information about $q$. Because $p_1<1/2$ for finite $V$, matching $15\%$ in expectation also forces $q>(0.15-p_0)/(0.5-p_0)\approx0.267$. Thus the summary suggests tension with the earlier prior concentrated near $0.1$, under this mean-zero alternative model. The event is the one-sided exceedance $Y_i>2$, not $|Y_i|>2$.
Back to article page
