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I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Also, be sure not to miss this related question already quoted by whuber in the comments.

A few hints for numerical optimization

In the above code the numerical optimization is done using nlm, a function of R. Here are a few hints for a more elementary solution.

If $b\ne 0$ is fixed, it is easy to find $a$ and $c$ minimizing $f(a,b,c) = \sum_{i=1}^n (a \exp(-b\cdot t_i) +c -y_i)^2$: this a the classical least squares for the linear model. Namely, letting $x_i = \exp(-b\cdot t_i)$, we find $$ \begin{array}{rcccl} a &=& a(b) &=& { n \sum_i y_i x_i - \left( \sum_i y_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }, \\ c &=& c(b) &=& { \sum_i y_i \sum_i x_i^2 - \left( \sum_i y_i x_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }. \\ \end{array}$$

Then set $$ g(b) = f(a(b), b, c(b)).$$ The optimization problem is now reduced to find the minimum of $g(b)$. This can be done either using Newton method or a ternary search.

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Also, be sure not to miss this related question already quoted by whuber in the comments.

A few hints for numerical optimization

In the above code the numerical optimization is done using nlm, a function of R. Here are a few hints for a more elementary solution.

If $b\ne 0$ is fixed, it is easy to find $a$ and $c$ minimizing $f(a,b,c) = \sum_{i=1}^n (a \exp(-b\cdot t_i) +c -y_i)^2$: this a the classical least squares for the linear model. Namely, letting $x_i = \exp(-b\cdot t_i)$, we find $$ \begin{array}{rcccl} a &=& a(b) &=& { n \sum_i y_i x_i - \left( \sum_i y_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }, \\ c &=& c(b) &=& { \sum_i y_i \sum_i x_i^2 - \left( \sum_i y_i x_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }. \\ \end{array}$$

Then set $$ g(b) = f(a(b), b, c(b)).$$ The optimization problem is now reduced to find the minimum of $g(b)$. This can be done either using Newton method or a ternary search.

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Also, be sure not to miss this related question already quoted by whuber in the comments.

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Also, be sure not to miss this related question already quoted by whuber in the comments.

A few hints for numerical optimization

In the above code the numerical optimization is done using nlm, a function of R. Here are a few hints for a more elementary solution.

If $b\ne 0$ is fixed, it is easy to find $a$ and $c$ minimizing $f(a,b,c) = \sum_{i=1}^n (a \exp(-b\cdot t_i) +c -y_i)^2$: this a the classical least squares for the linear model. Namely, letting $x_i = \exp(-b\cdot t_i)$, we find $$ \begin{array}{rcccl} a &=& a(b) &=& { n \sum_i y_i x_i - \left( \sum_i y_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }, \\ c &=& c(b) &=& { \sum_i y_i \sum_i x_i^2 - \left( \sum_i y_i x_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }. \\ \end{array}$$

Then set $$ g(b) = f(a(b), b, c(b)).$$ The optimization problem is now reduced to find the minimum of $g(b)$. This can be done either using Newton method or a ternary search.

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

Least square are easy with numerical methods, here is piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Also, be sure not to miss this related question already quoted by whuber in the comments.