An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with $\mu =4$and ${\sigma }^2=4$, whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters $(6,9)$. A point is randomly chosen on the image and has a reading of $5$. If the fraction of the image that is black is $\alpha$, for what value of $\alpha$would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region?

The image is partioned into two regions: White and Black. A point on the white region is randomly distributed normal variable with mean ${\mu }_1=4$and variance ${\sigma }_1^2=4$and a point in the black region is normally distributed with mean ${\mu }_2=6$and variance ${\sigma }_2^2=9$.

To find: The value of $\alpha$where when a point chosen randomly has value $5$and $\alpha$is the fraction of portion in the black region such that the probability of making an error be the same regardless of whether one concludes it is in black region or white.

Since it is given that the probability of making an error of concluding it in either black or white region is same, we can assume without loss of generality for it to be in black and we have

$P\left[\text{the given random point belongs to the black region}\right]=\frac{1}{2}$

The calculation will be,

$$\begin{gathered} P\left[\text{the given random point belongs to the black region}\right]=\frac{\alpha \times P[\text{value}=5\mid \text{black}]}{\alpha \times P[\text{value}=5\mid \text{black}]+(1-\alpha )\times P[\text{value}=5\mid \text{white}]} \\ =\frac{\alpha \times \frac{e^{-1/8}}{2\sqrt{2\pi }}}{\alpha \times \frac{e^{-1/8}}{2\sqrt{2\pi }}+(1-\alpha )\times \frac{e^{-1/18}}{3\sqrt{2\pi }}} \\ =\frac{\frac{\alpha }{2}}{\frac{\alpha }{2}+\frac{(1-\alpha )\times e^{-5/72}}{3}} \\ =\frac{3\alpha }{3\alpha +2(1-\alpha )\times e^{-5/72}} \end{gathered}$$

Here we need to substitute the value,

$$\begin{gathered} \frac{3\alpha }{3\alpha +2(1-\alpha )\times e^{-5/72}}=\frac{1}{2} \\ 6\alpha =3\alpha +2(1-\alpha )\times e^{-5/72} \\ 3\alpha =2(1-\alpha )\times 0.93 \\ 3\alpha =1.86-1.86\alpha \\ \alpha =0.3827 \end{gathered}$$