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For a normal distribution,

$$\phi(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\times e^{\frac{-1(x-\mu)^2}{2\sigma^2}}$$

Question:Prove that the standard normal distribution has 0 mean and unit variance?

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Lemmas we will use before we begin:

If $f(x) = -f(-x)$ then $\int_a^a f(x) dx = 0 $ for any $a \in \mathbb{R} $

Mean

$\mathbb{E}[X] = \int_{\mathbb{R}} xf_X(x) dx$

Using lemma above and noticing that $xf_X(x)$ satisfies the property above we are done for zero mean.

Variance

$\mathbb{V}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \mathbb{E}[X^2]$ as $\mathbb{E}[X] = 0 $

Hence $\mathbb{V}[X] = \int_{\mathbb{R}} x^2f_X(x) dx = 1$

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