Message Boards Message Boards

What is the interpretation of the 2 below plots for the given function ?

I have tried to plot that function which i have considred it to be a pdf which it is defined as : $$F(\sigma,\mu,z)=\frac{\pi ^2 (z-\sigma )^2 \exp \left(-\frac{(z-\sigma )^2 \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi } \text{erf}\left(\frac{(z-\sigma )^2 \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }}{1+0.25 \mu ^2}\right)}{1+0.25 \mu ^2}\right)}{0.961794 \left(\left(1+0.25 \mu ^2\right) \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }\right)}$$, such that : $\mu \in (0,1)$ and $\sigma ,z$ are reals. Now the integrand over sigma with sigma positive I got the first plot . Code of the plot 1:

Plot[NIntegrate[((Pi^2/0.961794)*(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2)*Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]) Exp[-(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2))*
  Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]*Erf[(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2)) Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]]])/.{\[Sigma]->-10},{z,0, Infinity}] ,{\[Mu],0,1}]

and the Code of the second Plot is :

Plot[NIntegrate[((Pi^2/0.961794)*(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2)*Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]) Exp[-(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2))*
  Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]*Erf[(z - \[Sigma])^2/((1 + 0.25 \[Mu]^2)) Sqrt[(1 + 0.25 \[Mu]^2)*2*Pi]]])/.{\[Sigma]->10},{z,0, Infinity}] ,{\[Mu],0,1}]

enter image description here enter image description here

Now my question here : Does those two plots means that the probability is 0 or 1 ?
POSTED BY: Rafik zeraoulia
2 Replies
Sort By:
Bill Nelson
Bill Nelson
Posted 4 years ago

If I plot your two integrands on the same sheet, each with your selected value for Sigma and both with the same plausible value of Mu and over a range large enough to capture and display the interesting part of both graphs and with PlotRange->All to display the full vertical extent of your functions:

Plot[{
  ((Pi^2/0.961794)*(z-\[Sigma])^2/((1+0.25 \[Mu]^2)*Sqrt[(1+0.25 \[Mu]^2)*2*Pi])*
    Exp[-(z-\[Sigma])^2/((1+0.25 \[Mu]^2))*Sqrt[(1+0.25 \[Mu]^2)*2*Pi]*Erf[(z-\[Sigma])^2/
    ((1+0.25 \[Mu]^2)) Sqrt[(1+0.25 \[Mu]^2)*2*Pi]]])/.{\[Sigma]->-10,\[Mu]->1},
  ((Pi^2/0.961794)*(z-\[Sigma])^2/((1+0.25 \[Mu]^2)*Sqrt[(1+0.25 \[Mu]^2)*2*Pi])*
    Exp[-(z-\[Sigma])^2/((1+0.25 \[Mu]^2))*Sqrt[(1+0.25 \[Mu]^2)*2*Pi]*Erf[(z-\[Sigma])^2/
    ((1+0.25 \[Mu]^2)) Sqrt[(1+0.25 \[Mu]^2)*2*Pi]]])/.{\[Sigma]->10,\[Mu]->1}},
  {z,-12,12},PlotRange->All]

then the interpretation is

Your Sigma->-10 shifted the curve off to the left of z==zero and thus your integral over z from zero to infinity was almost exactly zero

Your Sigma->10 shifted the curve off to the right of z==zero and thus your integral over z from 0 to infinity was substantially greater than zero

Now at a much higher level. I understand you are really struggling trying to find something. I see that the questions you are asking and the replies you are getting are not giving you what you need. Can you put all that aside for a few minutes and think at a very high level? Forget for a few moments Mean and Variance and which variable you are integrating over and what a Plot looks like and what Mathematica syntax is.

What is it that you are really really trying to get at the very highest level?

Perhaps when that is very clear it might be possible to move towards what you really need.
POSTED BY: Bill Nelson
Rohit Namjoshi
Rohit Namjoshi
Posted 4 years ago

Similar question posted here.
POSTED BY: Rohit Namjoshi
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract