Mixing
Log likelihood of normal time series
$begingroup$
Given a time series ${x_i, i=1, ldots, n}$ and define the corresponding series
$$y_i = log(x_{i+1} + a) - log(x_i + a),$$
where we assume that $y_i sim N(mu, sigma^2)$. The aim is to estimate $a$ via maximum likelihood estimation. I am coming across this in a financial paper and where the goal is to derive $a$ by MLE. The authors basically say the following:
For fixed $a$, it is easy to estimate $mu$ and $sigma$ since $y_i$ follows a normal distribution, hence this would be the sample mean and sample variance respectively.
Given $mu(a)$ and $sigma(a)$, one may determine $a$ as:
$$a := operatorname*{argmax}_a left(sum_{i=1}^{n} log f(x_{i+1}; x_i, a) right)$$
and they eventually derive the density $f$ using the transformation theorem:
begin{align}
f(x_{i+1};x_i,a) &= phi(x, mu(a), sigma(a))frac{d}{dx_{i+1}}|_{x = log(x_{i+1}+a)-log(x_i+a)} \
&= ... \
&= frac{1}{sqrt{2pi} sigma(a)(x_{i+1}+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{2sigma(a)^2} right }
end{align}
It has been some time that I have worked with log likelihoods so I am not sure if I can derive it correctly myself. First of all, we are dealing with a time series so I guess the assumption of independence is not really valid and instead we have to look at conditional densities.
So my reasoning was the following:
Since $y_i sim N(mu, sigma^2)$, we can say that $$y_i|x_{i} sim N(mu + log(x_{i}+a), sigma^2)$$
Now the authors derive the density in function of $y_i$ instead of $x_i$ so they use the transformation theorem to go back from $y$ to $x$. So let's define a random variable $Y = log(X+a)|{x_i}$ then the function $f: x to e^{x}-a$ maps $Y$ to $X$ with inverse $f^{-1}:x to log(x+a)$. Therefore by the transformation theorem the conditional density of $X|_{x_i}$ is then given by:
$$f_Y|_{x_i}(log(x+a))frac{d}{dx}log(x+a) = f_{Y|_{x_i}}(log(x+a))frac{1}{x+a}$$
and substituting the normal density I arrive to:
$$
frac{1}{sqrt{2pi}sigma(a) (x+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{sigma(a)^2} right }
$$
(I am not sure where the authors get the extra factor $2$ from in the denominator?)
First of all is my reasoning correct? Secondly, is the conditional density of each $x_{i+1}|x_i$ enough to get the total log-likelihood? Or is there a term missing?
Thanks!
maximum-likelihood log-likelihood
asked Jan 16 at 18:41
user39039user39039
808
$endgroup$
add a comment |
$begingroup$
Given a time series ${x_i, i=1, ldots, n}$ and define the corresponding series
$$y_i = log(x_{i+1} + a) - log(x_i + a),$$
where we assume that $y_i sim N(mu, sigma^2)$. The aim is to estimate $a$ via maximum likelihood estimation. I am coming across this in a financial paper and where the goal is to derive $a$ by MLE. The authors basically say the following:
For fixed $a$, it is easy to estimate $mu$ and $sigma$ since $y_i$ follows a normal distribution, hence this would be the sample mean and sample variance respectively.
Given $mu(a)$ and $sigma(a)$, one may determine $a$ as:
$$a := operatorname*{argmax}_a left(sum_{i=1}^{n} log f(x_{i+1}; x_i, a) right)$$
and they eventually derive the density $f$ using the transformation theorem:
begin{align}
f(x_{i+1};x_i,a) &= phi(x, mu(a), sigma(a))frac{d}{dx_{i+1}}|_{x = log(x_{i+1}+a)-log(x_i+a)} \
&= ... \
&= frac{1}{sqrt{2pi} sigma(a)(x_{i+1}+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{2sigma(a)^2} right }
end{align}
It has been some time that I have worked with log likelihoods so I am not sure if I can derive it correctly myself. First of all, we are dealing with a time series so I guess the assumption of independence is not really valid and instead we have to look at conditional densities.
So my reasoning was the following:
Since $y_i sim N(mu, sigma^2)$, we can say that $$y_i|x_{i} sim N(mu + log(x_{i}+a), sigma^2)$$
Now the authors derive the density in function of $y_i$ instead of $x_i$ so they use the transformation theorem to go back from $y$ to $x$. So let's define a random variable $Y = log(X+a)|{x_i}$ then the function $f: x to e^{x}-a$ maps $Y$ to $X$ with inverse $f^{-1}:x to log(x+a)$. Therefore by the transformation theorem the conditional density of $X|_{x_i}$ is then given by:
$$f_Y|_{x_i}(log(x+a))frac{d}{dx}log(x+a) = f_{Y|_{x_i}}(log(x+a))frac{1}{x+a}$$
and substituting the normal density I arrive to:
$$
frac{1}{sqrt{2pi}sigma(a) (x+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{sigma(a)^2} right }
$$
(I am not sure where the authors get the extra factor $2$ from in the denominator?)
First of all is my reasoning correct? Secondly, is the conditional density of each $x_{i+1}|x_i$ enough to get the total log-likelihood? Or is there a term missing?
Thanks!
maximum-likelihood log-likelihood
asked Jan 16 at 18:41
user39039user39039
808
$endgroup$
add a comment |
$begingroup$
Given a time series ${x_i, i=1, ldots, n}$ and define the corresponding series
$$y_i = log(x_{i+1} + a) - log(x_i + a),$$
where we assume that $y_i sim N(mu, sigma^2)$. The aim is to estimate $a$ via maximum likelihood estimation. I am coming across this in a financial paper and where the goal is to derive $a$ by MLE. The authors basically say the following:
For fixed $a$, it is easy to estimate $mu$ and $sigma$ since $y_i$ follows a normal distribution, hence this would be the sample mean and sample variance respectively.
Given $mu(a)$ and $sigma(a)$, one may determine $a$ as:
$$a := operatorname*{argmax}_a left(sum_{i=1}^{n} log f(x_{i+1}; x_i, a) right)$$
and they eventually derive the density $f$ using the transformation theorem:
begin{align}
f(x_{i+1};x_i,a) &= phi(x, mu(a), sigma(a))frac{d}{dx_{i+1}}|_{x = log(x_{i+1}+a)-log(x_i+a)} \
&= ... \
&= frac{1}{sqrt{2pi} sigma(a)(x_{i+1}+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{2sigma(a)^2} right }
end{align}
It has been some time that I have worked with log likelihoods so I am not sure if I can derive it correctly myself. First of all, we are dealing with a time series so I guess the assumption of independence is not really valid and instead we have to look at conditional densities.
So my reasoning was the following:
Since $y_i sim N(mu, sigma^2)$, we can say that $$y_i|x_{i} sim N(mu + log(x_{i}+a), sigma^2)$$
Now the authors derive the density in function of $y_i$ instead of $x_i$ so they use the transformation theorem to go back from $y$ to $x$. So let's define a random variable $Y = log(X+a)|{x_i}$ then the function $f: x to e^{x}-a$ maps $Y$ to $X$ with inverse $f^{-1}:x to log(x+a)$. Therefore by the transformation theorem the conditional density of $X|_{x_i}$ is then given by:
$$f_Y|_{x_i}(log(x+a))frac{d}{dx}log(x+a) = f_{Y|_{x_i}}(log(x+a))frac{1}{x+a}$$
and substituting the normal density I arrive to:
$$
frac{1}{sqrt{2pi}sigma(a) (x+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{sigma(a)^2} right }
$$
(I am not sure where the authors get the extra factor $2$ from in the denominator?)
First of all is my reasoning correct? Secondly, is the conditional density of each $x_{i+1}|x_i$ enough to get the total log-likelihood? Or is there a term missing?
Thanks!
maximum-likelihood log-likelihood
asked Jan 16 at 18:41
user39039user39039
808
$endgroup$
Given a time series ${x_i, i=1, ldots, n}$ and define the corresponding series
$$y_i = log(x_{i+1} + a) - log(x_i + a),$$
where we assume that $y_i sim N(mu, sigma^2)$. The aim is to estimate $a$ via maximum likelihood estimation. I am coming across this in a financial paper and where the goal is to derive $a$ by MLE. The authors basically say the following:
For fixed $a$, it is easy to estimate $mu$ and $sigma$ since $y_i$ follows a normal distribution, hence this would be the sample mean and sample variance respectively.
Given $mu(a)$ and $sigma(a)$, one may determine $a$ as:
$$a := operatorname*{argmax}_a left(sum_{i=1}^{n} log f(x_{i+1}; x_i, a) right)$$
and they eventually derive the density $f$ using the transformation theorem:
begin{align}
f(x_{i+1};x_i,a) &= phi(x, mu(a), sigma(a))frac{d}{dx_{i+1}}|_{x = log(x_{i+1}+a)-log(x_i+a)} \
&= ... \
&= frac{1}{sqrt{2pi} sigma(a)(x_{i+1}+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{2sigma(a)^2} right }
end{align}
It has been some time that I have worked with log likelihoods so I am not sure if I can derive it correctly myself. First of all, we are dealing with a time series so I guess the assumption of independence is not really valid and instead we have to look at conditional densities.
So my reasoning was the following:
Since $y_i sim N(mu, sigma^2)$, we can say that $$y_i|x_{i} sim N(mu + log(x_{i}+a), sigma^2)$$
Now the authors derive the density in function of $y_i$ instead of $x_i$ so they use the transformation theorem to go back from $y$ to $x$. So let's define a random variable $Y = log(X+a)|{x_i}$ then the function $f: x to e^{x}-a$ maps $Y$ to $X$ with inverse $f^{-1}:x to log(x+a)$. Therefore by the transformation theorem the conditional density of $X|_{x_i}$ is then given by:
$$f_Y|_{x_i}(log(x+a))frac{d}{dx}log(x+a) = f_{Y|_{x_i}}(log(x+a))frac{1}{x+a}$$
and substituting the normal density I arrive to:
$$
frac{1}{sqrt{2pi}sigma(a) (x+a)}exp left{frac{-1}{2}frac{[log(x_{i+1}+a)-log(x_i+a)-mu(a)]^2}{sigma(a)^2} right }
$$
(I am not sure where the authors get the extra factor $2$ from in the denominator?)
First of all is my reasoning correct? Secondly, is the conditional density of each $x_{i+1}|x_i$ enough to get the total log-likelihood? Or is there a term missing?
Thanks!
maximum-likelihood log-likelihood
maximum-likelihood log-likelihood
asked Jan 16 at 18:41
user39039user39039
808
asked Jan 16 at 18:41
user39039user39039
808
asked Jan 16 at 18:41
user39039user39039
808
asked Jan 16 at 18:41
user39039user39039
808
asked Jan 16 at 18:41
user39039user39039
808
808
add a comment |
add a comment |
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