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I am going to assume that you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you can write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.

I am going to assume that you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution:

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you can write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.

I am going to assume you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you can write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.

I am going to assume that you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you can write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.

I am going to assume you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.

I am going to assume you have observations from $Y$ because you are specifying distributional assumptions on $X$ and $Z$. Then $Y=Z-X$.

Given that $Y=Z-X$, we have that the density of $Y$ can be written as the convolution

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+x}{c}\right)\dfrac{1}{\sigma}f_X\left(\dfrac{x}{\sigma}\right)dx,$$

where $f_Z$ is the standard $\alpha -$stable density and $f_X$ represents the standard normal density. Note that, using a change of variable, we can rewrite this density as follows

$$f_Y(y)=\int_{-\infty}^{\infty}\dfrac{1}{c}f_Z\left(\dfrac{y+u\sigma}{c}\right)f_X\left(u\right)du = \int_{-\infty}^{\infty}\dfrac{1}{\sigma}f_X\left(\dfrac{cu-y}{\sigma}\right)f_Z\left(u\right)du,$$

It might be difficult (if feasible) to obtain this density in closed form but it can be approximated by simulating $(x_1,...,x_N)$ from a standard normal distribution and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{c}f_Z\left(\dfrac{y+x_j \sigma}{c}\right),$$

or by simulating from a standard $\alpha-$stable distribution $(z_1,...,z_N)$ and calculating the average

$$f_Y(y;c,\sigma)\approx \dfrac{1}{N}\sum_{j=1}^N \dfrac{1}{\sigma}f_X\left(\dfrac{c z_j -y}{\sigma}\right),$$

Note that the first approximation implies an easy simulation but a difficult evaluation of the density $f_Z$. The second approximation implies a difficult simulation but an easy evaluation of $f_X$. I also guess that the number of needed simulations for a good approximation is less in the first approximation. Either way, this might imply the use of intensive computation.

Now, if you have a sample $(y_1,...,y_n)$, then you can write the likelihood of $(c,\sigma)$ as

$${\mathcal L}(c,\sigma)\propto \prod_{j=1}^n f_Y(y_j;c,\sigma).$$

Using this, you can approximate the Maximum Likelihood Estimators (MLE) of $(c,\sigma)$ by maximising the corresponding likelihood using the approximations I described above.

Toy example in R

In this example, consider $(c,\sigma)=(0.5,1)$, $n=100$ and $N=1000$.

rm(list=ls())
library(stabledist)

# Values of the theoretical parameters
alpha0 = 1.75
sigma0 = 1
c0 = 0.5
set.seed(2)

 # Simulated sample
 y0 = rnorm(100) - rstable(n=100, alpha=alpha0 , beta=0, gamma = c0, delta = 0, pm = 0)

# A histogram of the sample
hist(y0)

# Second approximation of the density of Y
fy = function(y,c,sigma,ns){
z = rstable(n=ns, alpha=alpha0 , beta=0, gamma = 1, delta = 0, pm = 0)
return( mean(dnorm(c*z-y),mean=0,std=sigma ))
}

# -log likelihood
ll = function(par){
temp = rep(0,length(y0))
if(par[1]>0&par[2]>0){
for(j in 1:length(y0)) temp[j]=fy(y0[j],par[1],par[2],1000)
return(-sum(log(temp)))
}
else return(Inf)
}

# optimisation 
optim(c(0.5,1),ll,control = list(maxit=500))

The estimators I got are $(\hat c,\hat\sigma)=(0.657,0.942)$, sort of close to the theoretical values.

You can easily play with this code to obtain the corresponding estimators for your sample.

I hope this helps.