1. **Problem Statement:** Given independent random variables $X_i \sim N(0, i\sigma^2)$ for $i=1,2,\ldots,n$, with pdf $$f(x_i|\sigma^2) = \frac{1}{\sqrt{2\pi i \sigma^2}} \exp\left(-\frac{x_i^2}{2 i \sigma^2}\right),$$ prove or disprove that $f(x_i|\sigma^2)$ belongs to the regular 1-parameter exponential family. 2. **Recall the form of 1-parameter exponential family:** $$f(x|\theta) = g(x) \exp\{\theta t(x) - \psi(\theta)\}$$ where $\psi'(\theta) = E[t(X)]$ and $\psi''(\theta) = \mathrm{Var}[t(X)]$. 3. **Rewrite the pdf:** $$f(x_i|\sigma^2) = \frac{1}{\sqrt{2\pi i \sigma^2}} \exp\left(-\frac{x_i^2}{2 i \sigma^2}\right) = \frac{1}{\sqrt{2\pi i}} \sigma^{-1} \exp\left(-\frac{x_i^2}{2 i \sigma^2}\right).$$ 4. **Express in exponential family form:** Let $\theta = -\frac{1}{2 \sigma^2}$ (parameter), then $$f(x_i|\theta) = \frac{1}{\sqrt{2\pi i}} \exp\left(\theta \frac{x_i^2}{i} - \frac{1}{2} \ln(-\frac{1}{2\theta})\right).$$ More precisely, $$f(x_i|\theta) = g(x_i) \exp\{\theta t_i(x_i) - \psi_i(\theta)\}$$ where - $g(x_i) = \frac{1}{\sqrt{2\pi i}}$, - $t_i(x_i) = \frac{x_i^2}{i}$, - $\theta = -\frac{1}{2 \sigma^2}$, - $\psi_i(\theta) = -\frac{1}{2} \ln(-2\theta)$. 5. **Check derivatives:** $$\psi_i'(\theta) = \frac{1}{2\theta} = E[t_i(X_i)] = E\left(\frac{X_i^2}{i}\right) = \frac{E(X_i^2)}{i} = \frac{i \sigma^2}{i} = \sigma^2,$$ but since $\theta = -\frac{1}{2 \sigma^2}$, then $$\psi_i'(\theta) = \frac{1}{2\theta} = -\sigma^2,$$ which contradicts the positive expectation. 6. **Resolution:** Rewrite $\psi_i(\theta)$ correctly: Since $$f(x_i|\theta) = \frac{1}{\sqrt{2\pi i}} \exp\left(\theta \frac{x_i^2}{i} - \psi_i(\theta)\right),$$ and the normalizing constant must satisfy $$\exp(-\psi_i(\theta)) = \int \frac{1}{\sqrt{2\pi i}} \exp\left(\theta \frac{x_i^2}{i}\right) dx_i,$$ which is the moment generating function of $\frac{X_i^2}{i}$. Since $X_i \sim N(0, i \sigma^2)$, $\frac{X_i^2}{i} \sim \sigma^2 \chi_1^2$, so $$\psi_i(\theta) = -\frac{1}{2} \ln(-2\theta),$$ valid for $\theta < 0$. 7. **Conclusion:** The pdf $f(x_i|\sigma^2)$ can be expressed in the form $$f(x_i|\theta) = g(x_i) \exp\{\theta t_i(x_i) - \psi_i(\theta)\}$$ with - $g(x_i) = \frac{1}{\sqrt{2\pi i}}$, - $t_i(x_i) = \frac{x_i^2}{i}$, - $\theta = -\frac{1}{2 \sigma^2}$, - $\psi_i(\theta) = -\frac{1}{2} \ln(-2\theta)$, which satisfies the regular 1-parameter exponential family form. Hence, **$f(x_i|\sigma^2)$ belongs to the regular 1-parameter exponential family.**