Dtjytyk

Direct Experience of Meditation

Passing functions in C++

Did the new image of black hole confirm the general theory of relativity?

Jazz greats knew nothing of modes. Why are they used to improvise on standards?

Statistical model of ligand substitution

What is the largest species of polychaete?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

What's the point in a preamp?

Antler Helmet: Can it work?

Unable to start mainnet node docker container

How do I keep my slimes from escaping their pens?

Fishing simulator

What items from the Roman-age tech-level could be used to deter all creatures from entering a small area?

Is 1 ppb equal to 1 μg/kg?

Cold is to Refrigerator as warm is to?

Stopping real property loss from eroding embankment

How to say 'striped' in Latin

Estimated State payment too big --> money back; + 2018 Tax Reform

Mortgage adviser recommends a longer term than necessary combined with overpayments

Can a non-EU citizen traveling with me come with me through the EU passport line?

Was credit for the black hole image misattributed?

Can smartphones with the same camera sensor have different image quality?

How does modal jazz use chord progressions?

Need a suitable toxic chemical for a murder plot in my novel

sum of two independent exponential distributions?

0

$begingroup$

The sum of two independent exponential distributions begin{aligned}{Z=omega_{1}X_{1}+omega_{2}X_{2}}end{aligned} is as follows.
begin{aligned}f_{Z}(z)=&int _{0}^{z}omega_{1}omega_{2}f_{X_{1}}(x_{1})f_{X_{2}}(z-x_{1}),dx_{1}\=omega_{1}omega_{2}&lambda _{1}lambda _{2}exp[-lambda _{2}z]int _{0}^{z}exp[(lambda _{2}-lambda _{1})x_{1}],dx_{1}\=&{frac {omega_{1}omega_{2}lambda _{1}lambda _{2}}{lambda _{2}-lambda _{1}}}left(exp[-lambda _{1}z]-exp[-lambda _{2}z]right)end{aligned}

But the following example doesn't follow the above equation.
begin{aligned}omega_1=exp(-0.114989504)end{aligned}
begin{aligned}omega_2=exp(-7.5448246293224)end{aligned}
begin{aligned}lambda_1=1/(4.3258*10^{-08})end{aligned}
begin{aligned}lambda_1=1/(1.8737*10^{-03})end{aligned}

The result is compared with conv(fX1,fX2)*(x_interval) in Matlab.

Can you tell me what the problem is?

probability exponential-distribution

share|cite|improve this question

asked Mar 23 at 12:41

kwonhjkwonhj

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The convolution in Matlab is a summation, not an integral. My guess is that you reached limits of numerical integration.
  $endgroup$
  – P. Quinton
  Mar 23 at 12:47
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I reduced the x_interval for the convolution. And I solve that problem. Thank you so much.
  $endgroup$
  – kwonhj
  Mar 23 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not really about the interval, it's more about the fact that sometimes $sum_n f(nDelta) Delta$ may not converge to $int f(x) dx$ when $Delta$ goes to $0$
  $endgroup$
  – P. Quinton
  Mar 23 at 12:58
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The sum of two independent exponential distributions begin{aligned}{Z=omega_{1}X_{1}+omega_{2}X_{2}}end{aligned} is as follows.
begin{aligned}f_{Z}(z)=&int _{0}^{z}omega_{1}omega_{2}f_{X_{1}}(x_{1})f_{X_{2}}(z-x_{1}),dx_{1}\=omega_{1}omega_{2}&lambda _{1}lambda _{2}exp[-lambda _{2}z]int _{0}^{z}exp[(lambda _{2}-lambda _{1})x_{1}],dx_{1}\=&{frac {omega_{1}omega_{2}lambda _{1}lambda _{2}}{lambda _{2}-lambda _{1}}}left(exp[-lambda _{1}z]-exp[-lambda _{2}z]right)end{aligned}

But the following example doesn't follow the above equation.
begin{aligned}omega_1=exp(-0.114989504)end{aligned}
begin{aligned}omega_2=exp(-7.5448246293224)end{aligned}
begin{aligned}lambda_1=1/(4.3258*10^{-08})end{aligned}
begin{aligned}lambda_1=1/(1.8737*10^{-03})end{aligned}

The result is compared with conv(fX1,fX2)*(x_interval) in Matlab.

Can you tell me what the problem is?

probability exponential-distribution

share|cite|improve this question

asked Mar 23 at 12:41

kwonhjkwonhj

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The convolution in Matlab is a summation, not an integral. My guess is that you reached limits of numerical integration.
  $endgroup$
  – P. Quinton
  Mar 23 at 12:47
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I reduced the x_interval for the convolution. And I solve that problem. Thank you so much.
  $endgroup$
  – kwonhj
  Mar 23 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not really about the interval, it's more about the fact that sometimes $sum_n f(nDelta) Delta$ may not converge to $int f(x) dx$ when $Delta$ goes to $0$
  $endgroup$
  – P. Quinton
  Mar 23 at 12:58
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The sum of two independent exponential distributions begin{aligned}{Z=omega_{1}X_{1}+omega_{2}X_{2}}end{aligned} is as follows.
begin{aligned}f_{Z}(z)=&int _{0}^{z}omega_{1}omega_{2}f_{X_{1}}(x_{1})f_{X_{2}}(z-x_{1}),dx_{1}\=omega_{1}omega_{2}&lambda _{1}lambda _{2}exp[-lambda _{2}z]int _{0}^{z}exp[(lambda _{2}-lambda _{1})x_{1}],dx_{1}\=&{frac {omega_{1}omega_{2}lambda _{1}lambda _{2}}{lambda _{2}-lambda _{1}}}left(exp[-lambda _{1}z]-exp[-lambda _{2}z]right)end{aligned}

But the following example doesn't follow the above equation.
begin{aligned}omega_1=exp(-0.114989504)end{aligned}
begin{aligned}omega_2=exp(-7.5448246293224)end{aligned}
begin{aligned}lambda_1=1/(4.3258*10^{-08})end{aligned}
begin{aligned}lambda_1=1/(1.8737*10^{-03})end{aligned}

The result is compared with conv(fX1,fX2)*(x_interval) in Matlab.

Can you tell me what the problem is?

probability exponential-distribution

share|cite|improve this question

asked Mar 23 at 12:41

kwonhjkwonhj

11

$endgroup$

share|cite|improve this question

asked Mar 23 at 12:41

kwonhjkwonhj

11

share|cite|improve this question

asked Mar 23 at 12:41

kwonhjkwonhj

11

asked Mar 23 at 12:41

kwonhjkwonhj

11

asked Mar 23 at 12:41

kwonhjkwonhj

11