Standard Deviation of a Normally Distributed Random Variable with an Exponential PDF

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Homework Statement


A Normally distributed random variable with mean μ has a probability
density function given by

_ρ_.....*........((-ρ2(x-μ)2)/2δ)
√2∏δ|........e^

Homework Equations


Its standard deviation is given by: A)ρ2/δ B)δ/ρ C)√δ|/ρ D)ρ/√δ| E)√δ|/2ρ


The Attempt at a Solution



Now I know that the probability density function of a normal distribution is given by one over sigma * the square root of two pi, all times e to the power of negative (x minus mu) squared over 2 sigma squared.

The differences between this equation and the PDF for a normally distributed random variable then are minute and come to ro instead of 1 and the square of sigma instead of sigma in the outer part and ro squared as well as sigma not being squared in the denominator of the exponential.

What I don't know is how to parse out the standard deviation from all of this. Am I to integrate it? If not, which parts of the probability density function comprise its variance or standard deviation. Any guidance would be much appreciated.

Also, apologies for my shoddy depiction and description of these equations. A clearer version of the problem I'm working on can be found here: http://www.7citylearning.com/cqf/pdf/maths_test.pdf It is question 13.

I'll also attach the notes I'm working off of in trying to solve this query.
 

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Answers and Replies

  • #2
lanedance
Homework Helper
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Don't worry too much about the constant at the front, this is just to normalise the distribution (though it will be a useful check), the exponential gives it its shape.

As sigma has already be taken, to prevent confusion, lets call the standard deviation d


so you have
[tex] e^{-\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}}[/tex]

a normal distribution with standard deviation d, has
[tex] e^{-\frac{(x-\mu)^2}{2d^2}}[/tex]

equating them we have
[tex] \frac{(x-\mu)^2}{2d^2}=\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}[/tex]

once you solve for d, you can check it makes sense in the normalisation constant.
 
  • #3
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Sir, you are a God among men. The brilliance of your statement, "As sigma has already be taken, to prevent confusion, lets call the standard deviation d", is beyond measure. You have my eternal gratitude.
 
  • #4
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I find the answer to be the following:

d = [itex]\frac{\sqrt{σ}}{\gamma}[/itex]

Where d is the standard deviation of variable for the given pdf. Could someone confirm/correct my result, please?

-Mike
 
Last edited:

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