1. **Problem Statement:** We are given the probability density function (pdf) of the standard normal distribution: $$f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}, \quad x \in \mathbb{R}$$ We want to understand and verify the properties: - The total area under the curve is 1. - The expected value $\mathbb{E}[X]$ is 0. - The expected value of the square $\mathbb{E}[X^2]$ is 1. 2. **Formula and Important Rules:** - The pdf integrates to 1 over the entire real line: $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$ - The expected value (mean) is: $$\mathbb{E}[X] = \int_{-\infty}^{\infty} x f(x) \, dx$$ - The variance is: $$\mathrm{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$ For the standard normal, mean is 0 and variance is 1. 3. **Verification of total area:** The integral of the standard normal pdf over $\mathbb{R}$ is known to be 1 by definition of a probability density function. 4. **Calculation of $\mathbb{E}[X]$:** Since $f(x)$ is an even function and $x$ is odd, the product $x f(x)$ is an odd function. The integral of an odd function over symmetric limits is 0: $$\int_{-\infty}^{\infty} x f(x) \, dx = 0$$ 5. **Calculation of $\mathbb{E}[X^2]$:** We compute: $$\mathbb{E}[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx = \int_{-\infty}^{\infty} x^2 \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \, dx$$ This integral evaluates to 1, which is the variance of the standard normal distribution. **Final answers:** - Total area under $f(x)$ is 1. - $\mathbb{E}[X] = 0$ - $\mathbb{E}[X^2] = 1$