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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 givesFor $V>0$, threshold inference for a mixture proportion therefore givesThis 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 $K\sim \mathrm{Binomial}(N,(1-q)p_0+q{p}_1)$, 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 $p_0\approx 0.02275$. Matching $15\%$ exactly requires $p_1\ge 0.15$, equivalently $V\ge [2/{\Phi }^{-1}(0.85){]}^2-1\approx 2.724$. 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 $q>(0.15-p_0)/(0.5-p_0)\approx 0.267$. 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 $|Y_i|>2$.