- #1
cpwang
- 2
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Hi,
I am new here, so I have also attached my formula using pdf here.
I am trying to calculate the moments of a cut-off standard distribution. The question is pure calculus in nature. Please see the attached file.
I need to calculate:
\begin{equation}
M(j, \bar{R}) \equiv \frac{1}{\sqrt{2 \pi}} \int_{-\bar{R}}^{+\bar{R}} z^j e^{-\frac{z^2}{2}} dz;
\end{equation}
Here is how I proceeds:
\begin{align}
M(2j, \bar{R}) & = (2j-1)M(2j-2, \bar{R}) - \frac{2}{\sqrt{2\pi}} \bar{R}^{\frac{2j - 1}{2}} e^{-\frac{\bar{R}}{2}} \\
& = (2j - 1)! (M(0, \bar{R}) - \frac{2}{\sqrt{2\pi}} e^{-\frac{\bar{R}}{2}} \sum_{k=1}^{j} \frac{1}{(2k - 1)!} \bar{R}^{\frac{2k - 1}{2}}); \\
X \equiv & 2 e^{-\frac{\bar{R}}{2}} \sum_{k=1}^{j} \frac{1}{(2k - 1)!} \bar{R}^{\frac{2k - 1}{2}}:
\frac{d X}{d\bar{R}} = e^{-\frac{\bar{R}}{2}} {\bar{R}}^{-\frac{1}{2}} (1 - \frac{1}{(2j - 1)!} \bar{R}^{\frac{2j}{2}}); \\
z \equiv \sqrt{{\bar{R}}}: & X = \int \frac{d X}{d\bar{R}} d\bar{R}
= 2 \int_{0}^{\sqrt{{\bar{R}}}} e^{-\frac{z^2}{2}} (1 - \frac{1}{(2j - 1)!} z^{2j}) dz; \\
M(2j, \bar{R}) & = (2j - 1)! (M(0, \bar{R}) - M(0, \sqrt{{\bar{R}}})) + M(2j, \sqrt{{\bar{R}}}); \\
M(2j, \bar{R}) & - (2j - 1)! M(0, \bar{R}) = M(2j, \sqrt{{\bar{R}}}) - (2j - 1)! M(0, \sqrt{{\bar{R}}});
\end{align}
The last formula should no longer depends on $\bar{R}$. Because when $\bar{R}$ approaches $\infty$, $M(2j, \bar{R})$ approaches $(2j-1)!$ while $M(0, \bar{R})$ approaches 1, the conclusion of the above formula is:
\begin{equation}
\label{eqn: uncertainty moment 3}
M(2j, \bar{R}) = (2j - 1)! M(0, \bar{R});
\end{equation}
Unfortunately, this conclusion is empirically wrong. I did numerical calculation up to 200 terms, and found:
Where have I made an mistake?
This problem has troubled me for a few days, because I know the result is wrong, but could not find why.
I am new here, so I have also attached my formula using pdf here.
I am trying to calculate the moments of a cut-off standard distribution. The question is pure calculus in nature. Please see the attached file.
I need to calculate:
\begin{equation}
M(j, \bar{R}) \equiv \frac{1}{\sqrt{2 \pi}} \int_{-\bar{R}}^{+\bar{R}} z^j e^{-\frac{z^2}{2}} dz;
\end{equation}
Here is how I proceeds:
\begin{align}
M(2j, \bar{R}) & = (2j-1)M(2j-2, \bar{R}) - \frac{2}{\sqrt{2\pi}} \bar{R}^{\frac{2j - 1}{2}} e^{-\frac{\bar{R}}{2}} \\
& = (2j - 1)! (M(0, \bar{R}) - \frac{2}{\sqrt{2\pi}} e^{-\frac{\bar{R}}{2}} \sum_{k=1}^{j} \frac{1}{(2k - 1)!} \bar{R}^{\frac{2k - 1}{2}}); \\
X \equiv & 2 e^{-\frac{\bar{R}}{2}} \sum_{k=1}^{j} \frac{1}{(2k - 1)!} \bar{R}^{\frac{2k - 1}{2}}:
\frac{d X}{d\bar{R}} = e^{-\frac{\bar{R}}{2}} {\bar{R}}^{-\frac{1}{2}} (1 - \frac{1}{(2j - 1)!} \bar{R}^{\frac{2j}{2}}); \\
z \equiv \sqrt{{\bar{R}}}: & X = \int \frac{d X}{d\bar{R}} d\bar{R}
= 2 \int_{0}^{\sqrt{{\bar{R}}}} e^{-\frac{z^2}{2}} (1 - \frac{1}{(2j - 1)!} z^{2j}) dz; \\
M(2j, \bar{R}) & = (2j - 1)! (M(0, \bar{R}) - M(0, \sqrt{{\bar{R}}})) + M(2j, \sqrt{{\bar{R}}}); \\
M(2j, \bar{R}) & - (2j - 1)! M(0, \bar{R}) = M(2j, \sqrt{{\bar{R}}}) - (2j - 1)! M(0, \sqrt{{\bar{R}}});
\end{align}
The last formula should no longer depends on $\bar{R}$. Because when $\bar{R}$ approaches $\infty$, $M(2j, \bar{R})$ approaches $(2j-1)!$ while $M(0, \bar{R})$ approaches 1, the conclusion of the above formula is:
\begin{equation}
\label{eqn: uncertainty moment 3}
M(2j, \bar{R}) = (2j - 1)! M(0, \bar{R});
\end{equation}
Unfortunately, this conclusion is empirically wrong. I did numerical calculation up to 200 terms, and found:
- When $\bar{R}$ is below 1, $M(2j,\bar{R})$ decreases rapidly with j
- When $\bar{R}$ is below 6, $M(2j,\bar{R})$ increases exponentially with j
- When $\bar{R}$ is above 7, $M(2j,\bar{R})$ increases faster than exponentially with j
Where have I made an mistake?
This problem has troubled me for a few days, because I know the result is wrong, but could not find why.
Last edited:
