How does $f$ change with the increasing of $\sigma$ The $f=\int^{+\infty}_{0}x^{\lambda}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-m)^2}{2\sigma^2}}dx$, and $\lambda$ is a positive constant value, $m=1000$, how does $f$ change with the increasing of $\sigma$. (if $\sigma^2$ changes from 20 to 200)
How to prove it? Thanks.   
 A: If the integral were taken from $-\infty$ to $\infty$, then the integral equals to $\newcommand{\expect}{\mathbf{E}} \expect\left(X^\lambda\right)$ where $X\sim\mathcal{N}(m, \sigma^2)$.
So obviously when $\lambda=1$, it is nothing but the expected value of $X$, hence it stays the same, the value is the mean, $m$.
If $\lambda=2$, then it's $\newcommand{\var}{\mathbf{Var}}\expect X^2 = \var(X) + \left(\expect X\right)^2 = \var(X) + m^2$, hence it's increasing as $\sigma$ increases, since $\var(X) = \sigma^2$.
A: First note that
\begin{eqnarray*}
\newcommand{\iitt}[2]{\int_{#1}^{#2}}
f(\sigma) &=& \iitt{0}{\infty} x^\lambda \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x-m)^2}{2\sigma^2}
\right) dx\\
&=&\frac{1}{\sqrt{2\pi}} \iitt{-m/\sigma}{\infty} (\sigma y +m )^\lambda \exp(-y^2/2)dy.
\end{eqnarray*}
Now leibniz integral rule implies
\begin{eqnarray*}
f'(\sigma) &=& \frac{1}{\sqrt{2\pi}} \iitt{-m/\sigma}{\infty} \lambda (\sigma y + m)^{\lambda-1} \exp(-y^2/2) dy
- 0^\lambda \exp(-m^2/2\sigma^2) \cdot \frac{d}{d\sigma} (-m/\sigma)
\\
&=& \frac{1}{\sqrt{2\pi}} \iitt{-m/\sigma}{\infty} \lambda (\sigma y + m)^{\lambda-1} \exp(-y^2/2) dy > 0
\end{eqnarray*}
since $\lambda (\sigma y + m)^{\lambda-1} \exp(-y^2/2)>0$ for all $y>-m/\sigma$.
Therefore as long as $\lambda>0$, regardless of the values of $\lambda$ and $m$, $f(\sigma)$ is an increasing function of $\sigma$.
This proves that it is not true that $f(\sigma)$ is decreasing for $\lambda \in (0,1)$.
By the way, when $\lambda = 0$, it becomes
$$
f(\sigma) = \mathbf{Prob}\{X\geq -m/\sigma\}
$$
where $X\sim\mathcal{N}(0,1)$. Thus


*

*if $m=0$, $f(\sigma)$ is constant.

*if $m>0$, $-m/\sigma$ is an increasing function of $\sigma$, hence $f(\sigma)$ is a decreasing function of $\sigma$.

*if $m<0$, $-m/\sigma$ is a decreasing function of $\sigma$, hence $f(\sigma)$ is an increasing function of $\sigma$.

A: Taking the derivative
$$f'(\sigma) = \frac{1}{\sigma^4}\int_{0}^{\infty} \frac{x^\lambda}{\sqrt{2\pi}}((x-\mu)^2 - \sigma^2)e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$
Now let's take the case of $\mu=0$. This simplifies down to comparing the values of the two integrals 
$$I_1=\int_{0}^{\infty} x^{\lambda+2}e^{-\frac{x^2}{2\sigma^2}}dx \hspace{20 pt} I_2=\sigma^2\int_{0}^{\infty} x^{\lambda}e^{-\frac{x^2}{2\sigma^2}}dx$$
$$ I_1 = \sigma^{\lambda+3}2^{\frac{\lambda+1}{2}}\Gamma\left( \frac{\lambda+3}{2}\right) \hspace{20 pt} I_2 = \sigma^{\lambda+3}2^{\frac{\lambda-1}{2}}\Gamma\left( \frac{\lambda+1}{2}\right)$$
$$\implies \frac{I_1}{I_2} = \lambda + 1$$
In other words, this derivative is always positive for $\lambda > 0$, so it is always an increasing function of $\sigma$. I'm not sure yet how this changes in the $\mu\neq 0$ case.
