Question

Question:Suppose that a random variable X has a normal distribution for which the mean μ is unknown (−∞ <μ< ∞) and the variance σ2 is known. Let\({\bf{f}}\left( {{\bf{x}}\left| {\bf{\mu }} \right.} \right)\)denote the p.d.f. of X, and let\({\bf{f'}}\left( {{\bf{x}}\left| {\bf{\mu }} \right.} \right)\)and\({\bf{f''}}\left( {{\bf{x}}\left| {\bf{\mu }} \right.} \right)\)denote the first and second partial derivatives with respect to μ. Show that

\(\int_{{\bf{ - }}\infty }^\infty {{\bf{f'}}\left( {{\bf{x}}\left| {\bf{\mu }} \right.} \right)} {\bf{dx = 0}}\,\,{\bf{and}}\,\,\int_{{\bf{ - }}\infty }^\infty {{\bf{f''}}\left( {{\bf{x}}\left| {\bf{\mu }} \right.} \right)} {\bf{dx = 0}}{\bf{.}}\).

Short answer

Proved.

Step-by-step solution

Given information

X has a normal distribution with the mean of\(\mu \)and the variance of\({\sigma ^2}\).

\(f\left( {x\left| \mu \right.} \right)\)denote the p.d.f. of X

\(f'\left( {x\left| \mu \right.} \right)\) and \(f''\left( {x\left| \mu \right.} \right)\) denote the first and second-order partial derivatives with respect to \(\mu \)

Proving part

\(f\left( {x\left| \mu \right.} \right) = \frac{1}{{\sqrt {2\pi } \sigma }}\exp \left\{ { - \frac{1}{{2{\sigma ^2}}}{{\left( {x - \mu } \right)}^2}} \right\}\)

So,

The first-order derivative is,

\(\begin{array}{c}\frac{\partial }{{\partial \mu }}f\left( {x\left| \mu \right.} \right) = \frac{\partial }{{\partial \mu }}\left( {\frac{1}{{\sqrt {2\pi } \sigma }}\exp \left\{ { - \frac{1}{{2{\sigma ^2}}}{{\left( {x - \mu } \right)}^2}} \right\}} \right)\\ = \frac{1}{{\sqrt {2\pi } \sigma }}\frac{{\left( {x - \mu } \right)}}{{{\sigma ^2}}}\exp \left\{ { - \frac{1}{{2{\sigma ^2}}}{{\left( {x - \mu } \right)}^2}} \right\}\\ = \frac{{\left( {x - \mu } \right)}}{{{\sigma ^2}}}f\left( {x\left| \mu \right.} \right)\end{array}\)

The second-order derivative is,

\(\begin{array}{c}\frac{\partial }{{\partial \mu }}f'\left( {x\left| \mu \right.} \right) = \frac{\partial }{{\partial \mu }}\left( {\frac{{\left( {x - \mu } \right)}}{{{\sigma ^2}}}f\left( {x\left| \mu \right.} \right)} \right)\\ = \left[ {\frac{{{{\left( {x - \mu } \right)}^2}}}{{{\sigma ^4}}} - \frac{1}{{{\sigma ^2}}}} \right]f\left( {x\left| \mu \right.} \right)\end{array}\)

Therefore,

\(\begin{array}{c}\int_{ - \infty }^\infty {f'\left( {x\left| \mu \right.} \right)} dx = \frac{1}{{{\sigma ^2}}}\int_{ - \infty }^\infty {\left( {x - \mu } \right)} f\left( {x\left| \mu \right.} \right)d\mu \\ = \frac{1}{{{\sigma ^2}}}E\left( {X - \mu } \right)\\ = \frac{1}{{{\sigma ^2}}}\left( {\mu - \mu } \right)\\ = 0\end{array}\)

Hence, proved.

Again,

\(\begin{array}{c}\int_{ - \infty }^\infty {f''\left( {x\left| \mu \right.} \right)} dx = \frac{1}{{{\sigma ^4}}}E\left[ {{{\left( {X - \mu } \right)}^2}} \right] - \frac{1}{{{\sigma ^2}}}\\ = \frac{{{\sigma ^4}}}{{{\sigma ^2}}} - \frac{1}{{{\sigma ^2}}}\\ = \frac{1}{{{\sigma ^2}}} - \frac{1}{{{\sigma ^2}}}\\ = 0\end{array}\)

Hence, proved.