Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.
Sign up to join this community
I need to calculate the expected value of a random variable $X \sim N(1,3)$. The pdf for this variable is $\phi(x) = \frac{1}{\sqrt{6\pi}} exp\Big\{ - \frac{(x-1)^2}{6} \Big\}$.
The expected value is given as, \begin{align*} \therefore E(X) = \int_{-\infty}^{\infty} \frac{x}{\sqrt{6\pi}} exp\Big\{ - \frac{(x-1)^2}{6} \Big\} dx \end{align*}
I am stuck on the on the integral here. Am I correct so far?
[edit] Corrected the $\pi$ in denominator of the exponent.
The expected value of a random variable $X\sim\cal N\left( {1,3} \right)$ is 1.
However, as noted by Dilip Sarwate in his comment, your pdf is wrong: there should be no was wrong, there was an extra $\pi$ in the denominator of the exponent.
If you were looking for the calculations for the expected value of any Gaussian variable $X\sim\cal N\left( {\mu,\sigma^2} \right)$ $$E\left[ X \right] = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}\int\limits_{ - \infty }^\infty {x{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}dx} $$ one easy way is by substituting $z=x-\mu$, from which one obtains $$\begin{array}{} E\left[ X \right] &= \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}\int\limits_{ - \infty }^\infty {\left( {z + \mu } \right){e^{ - \frac{{{z^2}}}{{2{\sigma ^2}}}}}dz} \\ &= \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}\int\limits_{ - \infty }^\infty {z{e^{ - \frac{{{z^2}}}{{2{\sigma ^2}}}}}dz + \mu \left[ {\frac{1}{{\sqrt {2\pi {\sigma ^2}} }}\int\limits_{ - \infty }^\infty {{e^{ - \frac{{{z^2}}}{{2{\sigma ^2}}}}}dz} } \right]} \end{array}$$ The first integral evaluate to 0 because the integrand is an odd function and the integration can be split in two simmetric halves (with respect to the y axis), which are both convergent (i.e. the limits $\mathop {\lim }\limits_{a \to \infty } \int_0^a {f\left( x \right)dx} $ and $\mathop {\lim }\limits_{a \to -\infty } \int_a^0 {f\left( x \right)dx} $ that define them exist). The second integral (within square brackets) evaluates to 1 (it is a Gaussian pdf), so you are left with $$E\left[ X \right] = \mu$$
