1. **Problem statement:** Classify the point $x_0 = 0.3$ into one of three classes $Y=1,2,3$ based on posterior probabilities. 2. **Given:** - $Y=1 \sim N(\mu_1=0, \sigma^2=1)$ - $Y=2 \sim N(\mu_2=1, \sigma^2=1)$ - $Y=3$ is a mixture: $0.5 \times N(\mu_{31}=0.5, 1) + 0.5 \times N(\mu_{32}=0.5, 1)$ - Priors: $P(Y=1)=P(Y=2)=P(Y=3)=\frac{1}{3}$ 3. **Formula for posterior probability:** $$P(Y=k|x_0) = \frac{P(x_0|Y=k)P(Y=k)}{\sum_{j=1}^3 P(x_0|Y=j)P(Y=j)}$$ 4. **Likelihoods:** For normal distribution: $$P(x|Y=k) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x-\mu_k)^2}{2\sigma^2}\right)$$ 5. **Calculate likelihoods at $x_0=0.3$:** - $P(x_0|Y=1) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(0.3-0)^2}{2}\right) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{0.09}{2}\right)$ - $P(x_0|Y=2) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(0.3-1)^2}{2}\right) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{0.49}{2}\right)$ - $P(x_0|Y=3) = 0.5 \times \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(0.3-0.5)^2}{2}\right) + 0.5 \times \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(0.3-0.5)^2}{2}\right)$ Since both mixture components have the same mean 0.5, this simplifies to: $$P(x_0|Y=3) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(0.3-0.5)^2}{2}\right) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{0.04}{2}\right)$$ 6. **Calculate posterior probabilities (up to proportionality):** $$P(Y=k|x_0) \propto P(x_0|Y=k) P(Y=k) = \frac{1}{3} P(x_0|Y=k)$$ 7. **Compare posteriors:** - $P(Y=1|x_0) \propto \frac{1}{3\sqrt{2\pi}} \exp\left(-\frac{0.09}{2}\right)$ - $P(Y=2|x_0) \propto \frac{1}{3\sqrt{2\pi}} \exp\left(-\frac{0.49}{2}\right)$ - $P(Y=3|x_0) \propto \frac{1}{3\sqrt{2\pi}} \exp\left(-\frac{0.04}{2}\right)$ 8. **Decision:** Classify $x_0=0.3$ as the class with the highest posterior probability. Since $\exp\left(-\frac{0.04}{2}\right) > \exp\left(-\frac{0.09}{2}\right) > \exp\left(-\frac{0.49}{2}\right)$, the highest posterior is for class $Y=3$. **Final answer:** $$\boxed{\text{Classify } x_0=0.3 \text{ as } Y=3}$$