Change limits of integration.

[gino492]

I'm having problems changing the limits of integrations for the standard normal function.

The standard normal function has limits from negative infinity to x, and we know to change this we add a half to the function and make the new limits from 0 to x.

However now I want to change limits that are from positive infinity to x, but am confused on how to change these.

What do I do to the function in order to make this infinity a 0 or another constant?

Thanks

Gino

EDIT:
Here is a link to the picture of the integral of the standard normal function. The only difference for this question is that I have a positive infinity in the lower limit.
http://www.google.co.uk/imgres?q=in...224&start=0&ndsp=29&ved=1t:429,r:25,s:0,i:161

 

[DrPhil]

I'm having problems changing the limits of integrations for the standard normal function.

The standard normal function has limits from negative infinity to x, and we know to change this we add a half to the function and make the new limits from 0 to x.

However now I want to change limits that are from positive infinity to x, but am confused on how to change these.

What do I do to the function in order to make this infinity a 0 or another constant?

Thanks

Gino

Integrate as far as you have to go, and stop there. By its very nature, the normal distribution has a tail that extends to infinity. The formula for the unit normal itself is

\(\displaystyle \displaystyle f(x) = \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2}\)

The "cumulative" distribution, F(z), is the sum of all the area under the curve from -infinity to z:

\(\displaystyle \displaystyle F(z) = \int_{-\infty}^z \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2} dx\)

Since the function is symmetric about x=0, you can set the first half of the integral to exactly 1/2:

\(\displaystyle \displaystyle F(z) = \dfrac{1}{2} + \int_0^z \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2} dx\)

I haven't seen the file you mention in your EDIT.

 

[gino492]

Integrate as far as you have to go, and stop there. By its very nature, the normal distribution has a tail that extends to infinity. The formula for the unit normal itself is

\(\displaystyle \displaystyle f(x) = \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2}\)

The "cumulative" distribution, F(z), is the sum of all the area under the curve from -infinity to z:

\(\displaystyle \displaystyle F(z) = \int_{-\infty}^z \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2} dx\)

Since the function is symmetric about x=0, you can set the first half of the integral to exactly 1/2:

\(\displaystyle \displaystyle F(z) = \dfrac{1}{2} + \int_0^z \dfrac{1}{\sqrt{2 \pi}}\ e^{-x^2/2} dx\)

I haven't seen the file you mention in your EDIT.

Thanks Phil, but I'm trying to change the limit when it's a positive infinity not a negative.

 

[DrPhil]

Thanks Phil, but I'm trying to change the limit when it's a positive infinity not a negative.

This was not a change of limits, it was just using a constant for half of the domain. If you reverse the order of the limits, you make the function negative. What is it that you need??

 

[gino492]

This was not a change of limits, it was just using a constant for half of the domain. If you reverse the order of the limits, you make the function negative. What is it that you need??

So my limits for the function I want to integrate (I'm using a Monte Carlo hit and miss method on the R stats program)
are +infinity and x.

Now using the -infinity and x I know that you can just add the half to the function and use 0 and x.

What do I use when the limits are +infinity and x.

I want to get rid of the infinity and have a different limit, by using a similar method like using a constant for half the domain.

Does this help explain it better?

Gino.

 

[DrPhil]

So my limits for the function I want to integrate (I'm using a Monte Carlo hit and miss method on the R stats program)
are +infinity and x.

Now using the -infinity and x I know that you can just add the half to the function and use 0 and x.

What do I use when the limits are +infinity and x.

I want to get rid of the infinity and have a different limit, by using a similar method like using a constant for half the domain.

Does this help explain it better?

Gino.

The integral from +infinity to x is
the negative of the integral from x to +infinity,
which in turn is -1 + the integral from -infinity to x.

\(\displaystyle \displaystyle \int_{+\infty}^x f(t)\ dt = -\int_x^{+\infty} f(t)\ dt = -1+ \int_{-\infty}^x f(t)\ dt = -1+F(x) = -\dfrac{1}{2} + \int_0^x f(t)\ dt\)

 

[DrPhil]

I'm using a Monte Carlo hit and miss method on the R stats program

For a highly efficient Monte Carlo routine to select a random normal deviate, I recommend

Press et al. "Numerical Recipes, 2nd ed." (Cambridge,1992) section 7.2

The algorithm is essentially to find two uniform random deviates on (0,1), call then x_1 and x_2, and then convert them to two unit normal deviates y_1 and y_2 by the Box-Muller method:

\(\displaystyle y_1 = \sqrt{-2\ ln(x_1)}\cos (2\pi \ x_2) \)

\(\displaystyle y_2 = \sqrt{-2\ ln(x_1)}\sin (2\pi \ x_2) \)

The routine returns y_1 and saves y_2 for the next time it is called.