manoel_UFPE

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These are replies submitted by manoel_UFPE

Jakubi,

 

For what I understood the result of the integral goes to be

 

sum(2*(-1)^n*BesselK(-1/2+n, (1/2)*delta), n = 0 .. infinity)

 

it is this?

 

Manoel Neto

Federal University of Pernambuco

 

 

Jakubi,

 

For what I understood the result of the integral goes to be

 

sum(2*(-1)^n*BesselK(-1/2+n, (1/2)*delta), n = 0 .. infinity)

 

it is this?

 

Manoel Neto

Federal University of Pernambuco

 

 

Axel Vogt,

I understand what you want to say, however am wanting to calculate the value of integral to find the expected value of a probability distribution. In such a way the result to have that to be in function of the parameters of the distribution, that in this case are delta and mu. Therefore, it does not make sensible the numerical result of this integral.  Example: Either T a variate with distribution Birnbaum Saunders of parameters delta and mu. Soon its expected value is given by

E(T)=Int(t*g, t=0..infinity)

where

g:=exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

it unlaces way,

E(T)=Int(t*g,t=o..infinity)=(1/2)*(sqrt(mu*delta^2/(delta+1))*delta+delta*mu*sqrt((delta+1)/mu)+2*mu*sqrt((delta+1)/mu)+sqrt(mu*delta^2/(delta+1)))*exp(-(1/2)*sqrt((delta+1)/mu)*sqrt(mu*delta^2/(delta+1))+(1/2)*delta)*sqrt(Pi)*mu/(sqrt(Pi*delta*mu/(delta+1))*sqrt(1/delta)*(delta+1)^2) = mu.

To calculate the expected value of a function any, is enough to calculate the following integral

Int(f(t)*g,t=0..infinity)

in my problem the value of

f(t)=1/(t+delta*mu/(delta+1))^2.

Now I wait that they understand because it cannot be a numerical value.

 

 Manoel Neto

 

 

Axel Vogt,

I understand what you want to say, however am wanting to calculate the value of integral to find the expected value of a probability distribution. In such a way the result to have that to be in function of the parameters of the distribution, that in this case are delta and mu. Therefore, it does not make sensible the numerical result of this integral.  Example: Either T a variate with distribution Birnbaum Saunders of parameters delta and mu. Soon its expected value is given by

E(T)=Int(t*g, t=0..infinity)

where

g:=exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

it unlaces way,

E(T)=Int(t*g,t=o..infinity)=(1/2)*(sqrt(mu*delta^2/(delta+1))*delta+delta*mu*sqrt((delta+1)/mu)+2*mu*sqrt((delta+1)/mu)+sqrt(mu*delta^2/(delta+1)))*exp(-(1/2)*sqrt((delta+1)/mu)*sqrt(mu*delta^2/(delta+1))+(1/2)*delta)*sqrt(Pi)*mu/(sqrt(Pi*delta*mu/(delta+1))*sqrt(1/delta)*(delta+1)^2) = mu.

To calculate the expected value of a function any, is enough to calculate the following integral

Int(f(t)*g,t=0..infinity)

in my problem the value of

f(t)=1/(t+delta*mu/(delta+1))^2.

Now I wait that they understand because it cannot be a numerical value.

 

 Manoel Neto

 

 

In this case the parameters mu and delta are constant in the integral. Pi in the same way. I do not want numerical values of them, I only want a result in function of these constants.

 

I calculated this integral and gave certain. It looks at stops vocês understanding what desire to find as resulted.

 

int( (t+delta*mu/(delta+1))^(-1)*g , t = 0 .. infinity)

where

g := exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

Please they look at and they see if they understand my problem.

 

Manoel Neto

Federal University of Pernambuco

 

In this case the parameters mu and delta are constant in the integral. Pi in the same way. I do not want numerical values of them, I only want a result in function of these constants.

 

I calculated this integral and gave certain. It looks at stops vocês understanding what desire to find as resulted.

 

int( (t+delta*mu/(delta+1))^(-1)*g , t = 0 .. infinity)

where

g := exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

Please they look at and they see if they understand my problem.

 

Manoel Neto

Federal University of Pernambuco

 

The integral is correct, therefore I calculated other hopes with the same function and it did not have problem. However when I tried to calculate the expected value of the function below I did not get no result.

The function probability density is given by

f := exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

I am trying to calculate the expected value of the following function

g:=1/(t+delta*mu/(delta+1))^2

that is, that to calculate the following integral

int(g*f, t=0..infinity)

Manoel

Federal University of Pernambuco

The integral is correct, therefore I calculated other hopes with the same function and it did not have problem. However when I tried to calculate the expected value of the function below I did not get no result.

The function probability density is given by

f := exp(1/sqrt(2/delta)^2)*(t+delta*mu/(delta+1))*exp(-(t*(delta+1)/(delta*mu)+delta*mu/((delta+1)*t))/(2*sqrt(2/delta)^2))/((2*sqrt(2/delta)*sqrt(2*pi*delta*mu/(delta+1)))*t^(3/2))

I am trying to calculate the expected value of the following function

g:=1/(t+delta*mu/(delta+1))^2

that is, that to calculate the following integral

int(g*f, t=0..infinity)

Manoel

Federal University of Pernambuco

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