Mixing
Quadratic expansion of log-likelihood
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From the paper I'm reading:
Let $mathbf{x}$ represent the distribution of linear attenuation coefficient of an object and $[mathbf{Hx}]_m$ represents their line integral. The $m$th CT measurement, $y_m$, is a Poisson random variable with parameters
$$p_m sim Poisson(b_m e^{[mathbf{Hx}]_m}+r_m)$$
$$y_m = -log(frac{p_m}{b_m})$$
where $b_m$ is the blank scan factor and $r_m$ is the readout-noise. [Since the logarithm is bijective, the negative log-likelihood of $mathbf{y}$ given $mathbf{x}$ is equal to the one of $mathbf{p}$ given $mathbf{x}$. After removing the constants, we use the negative log-likelihood as the data-fidelity term
$$E(mathbf{Hx,y})= sum_{m=1}^M(hat{p}_m-p_m log(hat{p}_m))$$
where $hat{p}_m=b_m e^{-mathbf{Hx}}+r_m$ is the expected value of $p_m$. We then perform a quadratic approximation of $E$ with respect to $mathbf{Hx}$ around the point $-(ln(frac{hat{p}_m-r_m}{b_m}))$ using a Taylor expansion. After ignoring the higher-order terms, this yields
$$E(mathbf{Hx,y})=sum_{m=1}^M frac{1}{2} frac{(p_m-r_m)^2}{p_m}(mathbf{Hx}-log(frac{b_m}{p_m-r_m}))^2$$.
But I don't really know how to interpret the random variable $p_m$; can I just set $p_m=hat{p}_m$? That's what I've tried so far: let's assume $r_m=0$. Since $hat{p}_m=b_me^{-[mathbf{Hx}]_m}$, $-(ln(frac{hat{p}_m}{b_m}))$ becomes $[mathbf{Hx}]_m$, we can write the quadratic expansion as:
$$Q(E)=sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) \
+frac{d}{d mathbf{Hx}} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{1}{2} frac{d^2}{d mathbf{Hx}^2} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2$$
Furthermore:
$$E([mathbf{Hx}]_m,mathbf{y})=sum_{m=1}^M(b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}))$$
and
$$frac{d}{d mathbf{Hx}} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})$$
and
$$frac{d^2}{d mathbf{Hx}^2} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} (log(b_m e^{-[mathbf{Hx}]_m})+1)$$
Putting everything together and splitting the third term, I get:
$$Q(E)=sum_{m=1}^M (b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}) \
+b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})(mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (log(b_m e^{-[mathbf{Hx}]_m}))(mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2 \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2)$$
So, assuming $p_m=hat{p}_m$ the 4th term actually equals the proposed quadratic expansion from the paper; but is it correct what I did so far? And if it's right, how do I see that the first three terms vanish? Is there a trick or do I have to look closer?
Any kind of help is highly appreciated. Thank you very much. Btw. the paper I'm reading is: click (p. 1451, part B).
taylor-expansion log-likelihood
asked Jan 6 at 1:24
M. EvansM. Evans
257
$endgroup$
add a comment |
$begingroup$
From the paper I'm reading:
Let $mathbf{x}$ represent the distribution of linear attenuation coefficient of an object and $[mathbf{Hx}]_m$ represents their line integral. The $m$th CT measurement, $y_m$, is a Poisson random variable with parameters
$$p_m sim Poisson(b_m e^{[mathbf{Hx}]_m}+r_m)$$
$$y_m = -log(frac{p_m}{b_m})$$
where $b_m$ is the blank scan factor and $r_m$ is the readout-noise. [Since the logarithm is bijective, the negative log-likelihood of $mathbf{y}$ given $mathbf{x}$ is equal to the one of $mathbf{p}$ given $mathbf{x}$. After removing the constants, we use the negative log-likelihood as the data-fidelity term
$$E(mathbf{Hx,y})= sum_{m=1}^M(hat{p}_m-p_m log(hat{p}_m))$$
where $hat{p}_m=b_m e^{-mathbf{Hx}}+r_m$ is the expected value of $p_m$. We then perform a quadratic approximation of $E$ with respect to $mathbf{Hx}$ around the point $-(ln(frac{hat{p}_m-r_m}{b_m}))$ using a Taylor expansion. After ignoring the higher-order terms, this yields
$$E(mathbf{Hx,y})=sum_{m=1}^M frac{1}{2} frac{(p_m-r_m)^2}{p_m}(mathbf{Hx}-log(frac{b_m}{p_m-r_m}))^2$$.
But I don't really know how to interpret the random variable $p_m$; can I just set $p_m=hat{p}_m$? That's what I've tried so far: let's assume $r_m=0$. Since $hat{p}_m=b_me^{-[mathbf{Hx}]_m}$, $-(ln(frac{hat{p}_m}{b_m}))$ becomes $[mathbf{Hx}]_m$, we can write the quadratic expansion as:
$$Q(E)=sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) \
+frac{d}{d mathbf{Hx}} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{1}{2} frac{d^2}{d mathbf{Hx}^2} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2$$
Furthermore:
$$E([mathbf{Hx}]_m,mathbf{y})=sum_{m=1}^M(b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}))$$
and
$$frac{d}{d mathbf{Hx}} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})$$
and
$$frac{d^2}{d mathbf{Hx}^2} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} (log(b_m e^{-[mathbf{Hx}]_m})+1)$$
Putting everything together and splitting the third term, I get:
$$Q(E)=sum_{m=1}^M (b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}) \
+b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})(mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (log(b_m e^{-[mathbf{Hx}]_m}))(mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2 \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2)$$
So, assuming $p_m=hat{p}_m$ the 4th term actually equals the proposed quadratic expansion from the paper; but is it correct what I did so far? And if it's right, how do I see that the first three terms vanish? Is there a trick or do I have to look closer?
Any kind of help is highly appreciated. Thank you very much. Btw. the paper I'm reading is: click (p. 1451, part B).
taylor-expansion log-likelihood
asked Jan 6 at 1:24
M. EvansM. Evans
257
$endgroup$
add a comment |
$begingroup$
From the paper I'm reading:
Let $mathbf{x}$ represent the distribution of linear attenuation coefficient of an object and $[mathbf{Hx}]_m$ represents their line integral. The $m$th CT measurement, $y_m$, is a Poisson random variable with parameters
$$p_m sim Poisson(b_m e^{[mathbf{Hx}]_m}+r_m)$$
$$y_m = -log(frac{p_m}{b_m})$$
where $b_m$ is the blank scan factor and $r_m$ is the readout-noise. [Since the logarithm is bijective, the negative log-likelihood of $mathbf{y}$ given $mathbf{x}$ is equal to the one of $mathbf{p}$ given $mathbf{x}$. After removing the constants, we use the negative log-likelihood as the data-fidelity term
$$E(mathbf{Hx,y})= sum_{m=1}^M(hat{p}_m-p_m log(hat{p}_m))$$
where $hat{p}_m=b_m e^{-mathbf{Hx}}+r_m$ is the expected value of $p_m$. We then perform a quadratic approximation of $E$ with respect to $mathbf{Hx}$ around the point $-(ln(frac{hat{p}_m-r_m}{b_m}))$ using a Taylor expansion. After ignoring the higher-order terms, this yields
$$E(mathbf{Hx,y})=sum_{m=1}^M frac{1}{2} frac{(p_m-r_m)^2}{p_m}(mathbf{Hx}-log(frac{b_m}{p_m-r_m}))^2$$.
But I don't really know how to interpret the random variable $p_m$; can I just set $p_m=hat{p}_m$? That's what I've tried so far: let's assume $r_m=0$. Since $hat{p}_m=b_me^{-[mathbf{Hx}]_m}$, $-(ln(frac{hat{p}_m}{b_m}))$ becomes $[mathbf{Hx}]_m$, we can write the quadratic expansion as:
$$Q(E)=sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) \
+frac{d}{d mathbf{Hx}} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{1}{2} frac{d^2}{d mathbf{Hx}^2} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2$$
Furthermore:
$$E([mathbf{Hx}]_m,mathbf{y})=sum_{m=1}^M(b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}))$$
and
$$frac{d}{d mathbf{Hx}} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})$$
and
$$frac{d^2}{d mathbf{Hx}^2} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} (log(b_m e^{-[mathbf{Hx}]_m})+1)$$
Putting everything together and splitting the third term, I get:
$$Q(E)=sum_{m=1}^M (b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}) \
+b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})(mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (log(b_m e^{-[mathbf{Hx}]_m}))(mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2 \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2)$$
So, assuming $p_m=hat{p}_m$ the 4th term actually equals the proposed quadratic expansion from the paper; but is it correct what I did so far? And if it's right, how do I see that the first three terms vanish? Is there a trick or do I have to look closer?
Any kind of help is highly appreciated. Thank you very much. Btw. the paper I'm reading is: click (p. 1451, part B).
taylor-expansion log-likelihood
asked Jan 6 at 1:24
M. EvansM. Evans
257
$endgroup$
From the paper I'm reading:
Let $mathbf{x}$ represent the distribution of linear attenuation coefficient of an object and $[mathbf{Hx}]_m$ represents their line integral. The $m$th CT measurement, $y_m$, is a Poisson random variable with parameters
$$p_m sim Poisson(b_m e^{[mathbf{Hx}]_m}+r_m)$$
$$y_m = -log(frac{p_m}{b_m})$$
where $b_m$ is the blank scan factor and $r_m$ is the readout-noise. [Since the logarithm is bijective, the negative log-likelihood of $mathbf{y}$ given $mathbf{x}$ is equal to the one of $mathbf{p}$ given $mathbf{x}$. After removing the constants, we use the negative log-likelihood as the data-fidelity term
$$E(mathbf{Hx,y})= sum_{m=1}^M(hat{p}_m-p_m log(hat{p}_m))$$
where $hat{p}_m=b_m e^{-mathbf{Hx}}+r_m$ is the expected value of $p_m$. We then perform a quadratic approximation of $E$ with respect to $mathbf{Hx}$ around the point $-(ln(frac{hat{p}_m-r_m}{b_m}))$ using a Taylor expansion. After ignoring the higher-order terms, this yields
$$E(mathbf{Hx,y})=sum_{m=1}^M frac{1}{2} frac{(p_m-r_m)^2}{p_m}(mathbf{Hx}-log(frac{b_m}{p_m-r_m}))^2$$.
But I don't really know how to interpret the random variable $p_m$; can I just set $p_m=hat{p}_m$? That's what I've tried so far: let's assume $r_m=0$. Since $hat{p}_m=b_me^{-[mathbf{Hx}]_m}$, $-(ln(frac{hat{p}_m}{b_m}))$ becomes $[mathbf{Hx}]_m$, we can write the quadratic expansion as:
$$Q(E)=sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) \
+frac{d}{d mathbf{Hx}} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{1}{2} frac{d^2}{d mathbf{Hx}^2} left( sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y}) right) (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2$$
Furthermore:
$$E([mathbf{Hx}]_m,mathbf{y})=sum_{m=1}^M(b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}))$$
and
$$frac{d}{d mathbf{Hx}} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})$$
and
$$frac{d^2}{d mathbf{Hx}^2} sum_{m=1}^M E([mathbf{Hx}]_m,mathbf{y})= sum_{m=1}^M b_m e^{-[mathbf{Hx}]_m} (log(b_m e^{-[mathbf{Hx}]_m})+1)$$
Putting everything together and splitting the third term, I get:
$$Q(E)=sum_{m=1}^M (b_me^{-[mathbf{Hx}]_m}-b_me^{-[mathbf{Hx}]_m} log(b_me^{-[mathbf{Hx}]_m}) \
+b_m e^{-[mathbf{Hx}]_m} log(b_m e^{-[mathbf{Hx}]_m})(mathbf{Hx}+ln(frac{hat{p}_m}{b_m})) \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (log(b_m e^{-[mathbf{Hx}]_m}))(mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2 \
+frac{b_m e^{-[mathbf{Hx}]_m}}{2} (mathbf{Hx}+ln(frac{hat{p}_m}{b_m}))^2)$$
So, assuming $p_m=hat{p}_m$ the 4th term actually equals the proposed quadratic expansion from the paper; but is it correct what I did so far? And if it's right, how do I see that the first three terms vanish? Is there a trick or do I have to look closer?
Any kind of help is highly appreciated. Thank you very much. Btw. the paper I'm reading is: click (p. 1451, part B).
taylor-expansion log-likelihood
taylor-expansion log-likelihood
asked Jan 6 at 1:24
M. EvansM. Evans
257
asked Jan 6 at 1:24
M. EvansM. Evans
257
edited Jan 6 at 14:07
M. Evans
asked Jan 6 at 1:24
M. EvansM. Evans
257
asked Jan 6 at 1:24
M. EvansM. Evans
257
asked Jan 6 at 1:24
M. EvansM. Evans
257
257
add a comment |
add a comment |
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