how to find the integral of a generalization of the Gaussian: exp(-abs(x)^q/2)
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I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.
The problem is to prove that:
$$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
is normalized. I know how to take care of the absolute value of course:
$$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$
but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...
calculus integration statistics
edited Jan 18 at 22:22
idriskameni
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
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add a comment |
$begingroup$
I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.
The problem is to prove that:
$$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
is normalized. I know how to take care of the absolute value of course:
$$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$
but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...
calculus integration statistics
edited Jan 18 at 22:22
idriskameni
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
$endgroup$
add a comment |
$begingroup$
I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.
The problem is to prove that:
$$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
is normalized. I know how to take care of the absolute value of course:
$$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$
but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...
calculus integration statistics
edited Jan 18 at 22:22
idriskameni
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
$endgroup$
I am currently self studying through Bishop's pattern recognition and machine learning. I've hit a bit of a road block on exercise 2.43, and I'm hoping someone will be able to assist.
The problem is to prove that:
$$p(xlvert sigma^2, q) = frac{q}{2(2sigma^2)^{1/q}Gamma(1/q)}e^{-frac{lvert x rvert^q}{2sigma^2}}$$
is normalized. I know how to take care of the absolute value of course:
$$2int_0^infty e^{-frac{x^q}{2sigma^2}}$$
but... man. I apparently need to review my calc. I remember seeing a name for this kind of integral, but it was maybe a year ago, and I can't seem to find it again. Any help would be greatly appreciated. This dang question only has a one star difficulty, so maybe it's something stupid I'm just missing...
calculus integration statistics
calculus integration statistics
edited Jan 18 at 22:22
idriskameni
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
edited Jan 18 at 22:22
idriskameni
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
edited Jan 18 at 22:22
idriskameni
749321
edited Jan 18 at 22:22
idriskameni
749321
edited Jan 18 at 22:22
idriskameni
749321
749321
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
asked Jan 18 at 21:49
adventuringrawadventuringraw
103
103
add a comment |
add a comment |
1 Answer
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Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
$$
frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
$$
To finish off, use the definition of the Gamma function:
$$
Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
$$
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
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$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
$$
frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
$$
To finish off, use the definition of the Gamma function:
$$
Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
$$
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
$endgroup$
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
add a comment |
$begingroup$
Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
$$
frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
$$
To finish off, use the definition of the Gamma function:
$$
Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
$$
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
$endgroup$
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
add a comment |
$begingroup$
Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
$$
frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
$$
To finish off, use the definition of the Gamma function:
$$
Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
$$
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
$endgroup$
Clean up the exponent on $e$ by using the change of variables $t:=x^q/(2sigma^2)$, i.e., $x=(2sigma^2t)^{1/q}$, so that $dx=frac1q(2sigma^2t)^{frac1q-1}(2sigma^2)dt$. This transforms your integral into
$$
frac2q(2sigma^2)^{1/q}int_0^infty e^{-t}t^{frac1q-1},dt.
$$
To finish off, use the definition of the Gamma function:
$$
Gamma(z):=int_0^infty t^{z-1}e^{-t},dt
$$
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
answered Jan 18 at 22:28
grand_chatgrand_chat
20.5k11327
20.5k11327
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
add a comment |
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
$begingroup$
well shit. Thanks friend, I should probably sleep on problems before hitting stack exchange. I appreciate the help.
$endgroup$
– adventuringraw
Jan 18 at 23:30
add a comment |
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