Exponentially Modified Gaussian

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In summary, the conversation discusses fitting a detector response using Matlab and using an Exponentially Modified Gaussian. The equations for h(t) and f(t) are given, and the convolution y(t) is used to find the final equation y(t). There is also mention of completing the square for the terms in the exponent to find the z for the erf(z). However, the speaker is still working on completing the square correctly.
  • #1
James_1978
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I am fitting a detector response using Matlab. I have been asked to fit the spectrum using a Exponentially Modified Gaussian. This is as follows.

[ tex ] a^x_n [ /tex ]

h(t) = [ tex ] \frac{A}{\sqrt{2 \pi} \sigma} e^{-\frac{(t-t_{R})^{2}}{2\sigma^{2}} [ /tex ]

f(t) = [ tex ] \frac{1}{\tau}e^{-\frac{t}{tau}}[ /tex ]

Using the convolution y(t) = [ tex ] \int^{\infinity}_{0} h(t^{\prime})f(t-t^{prime})dt^{\prime}[ /tex ]

This gives y(t) = [ tex ] \frac{A}{\tau \sigma \sqrt{2 \pi}} \int^{\infinity}_{0}} e^{-{\frac{(t-t_{R} - t^{\prime})^{2}}{2\sigma^{2}}e^{\frac{-t^{\prime}}{tau}} [ /tex ]

This is the convolution of h(t)*f(t) to give y(t)

The answer is

y(t) = [ tex ] \frac{A}{2 \tau}[1-erf(\frac{\sigma}{\sqrt{2}\tau} - \frac{t-t_{R}}{\sqrt{2}\sigma]e^{\frac{\sigma^{2)}{\sqrt{2}\tau} - \frac{t-t_{R}}{\tau}}[ /tex ]

I have been unable to get the answer. Has anyone ever worked this out to get y(t)? Also this is my first post and I am not sure how to get the latex to appear?
 
Last edited:
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  • #2
James_1978 said:
I am fitting a detector response using Matlab. I have been asked to fit the spectrum using a Exponentially Modified Gaussian. This is as follows.

[tex]h(t) = \frac{A}{\sqrt{2 \pi} \sigma} e^{-\frac{(t-t_{R})^{2}}{2\sigma^{2}}[/tex]

[tex]f(t) = \frac{1}{\tau}e^{\frac{t}{\tau}}[/tex]

Using the convolution [tex]y(t) = \int^{\infinity}_{0} h(t^{\prime})f(t-t^{\prime})dt^{\prime}[/tex]

This gives [tex]y(t) = \frac{A}{\tau \sigma \sqrt{2 \pi}} \int^{\infinity}_{0} e^{-{-\frac{(t-t_{R} - t^{\prime})^{2}}{2\sigma^{2}}e^{\frac{t^{\prime}}{\tau}}[/tex]

This is the convolution of h(t)*f(t) to give f(t)

The answer is

[tex]y(t) = \frac{A}{2 \tau}[1-erf(\frac{\sigma}{\sqrt{2}\tau} - \frac{t-t_{R}}{\sqrt{2}\sigma})]e^{\frac{\sigma^{2}}{\sqrt{2}\tau} - \frac{t-t_{R}}{\tau}}[/tex]

I have been unable to get the answer. Has anyone ever worked this out to get y(t)? Also this is my first post and I am not sure how to get the latex to appear?

You must write your equations between [tex][tex][/tex][/tex]
 
Last edited:
  • #3
You need to prescribe the limits of integration on t'.
 
  • #4
Yes...but I believe one must complete the square for the terms in the exponent. This gives you the z for the erf(z). However, I am unable to complete the square correctly. I am close and still working on it.
 

1. What is an Exponentially Modified Gaussian (EMG) distribution?

The Exponentially Modified Gaussian (EMG) distribution is a probability distribution that combines the properties of the Gaussian (normal) distribution and the exponential distribution. It is commonly used to model data with a skewed shape, such as biological or financial data. The EMG distribution has three parameters: location, scale, and shape, which determine the mean, standard deviation, and skewness of the distribution, respectively.

2. How is the EMG distribution different from a normal distribution?

The EMG distribution differs from a normal distribution in that it has an additional parameter (shape) that allows for skewness, while the normal distribution assumes a symmetrical shape. The EMG distribution also has a longer tail on one side, depending on the value of the shape parameter.

3. What is the relationship between the EMG distribution and the Laplace distribution?

The Laplace distribution is a special case of the EMG distribution, where the shape parameter is set to 1. This results in a distribution that is symmetrical and has a sharp peak at the mean, with a heavier tail compared to the normal distribution. In contrast, the EMG distribution allows for more flexibility in modeling data with varying degrees of skewness.

4. How is the EMG distribution used in data analysis?

The EMG distribution can be used in data analysis to model skewed data and make predictions about future values. It is also commonly used in signal processing and image analysis, as well as in finance and economics to model stock prices and other financial data.

5. What are the advantages and limitations of using the EMG distribution?

The main advantage of using the EMG distribution is its flexibility in modeling skewed data, which cannot be accurately represented by the normal distribution. It also has a closed-form solution for its probability density function, making it easier to work with mathematically. However, the EMG distribution may not be suitable for all types of data and may require a larger sample size to accurately estimate its parameters compared to other distributions. Additionally, interpreting the shape parameter can be challenging and may require expert knowledge in the field of application.

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