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Computation of the moment of order 3 of Gamma distribution
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https://i.gyazo.com/9ab33847b7a2f8f7378f12e1ab5960b9.png
A continuous random variable $X$ has the $mathsfGamma(alpha, lambda)$ distribution with probability density
$$f_X(x) = fraclambda^alphaGamma(alpha)x^alpha - 1e^-lambda x,
text for x > 0.$$
Prove from first principles that $E(X^3) = fracalpha(alpha + 1)(alpha + 2)lambda^3.$
I don't know what it means when it says from first principles. Any help? Thanks.
statistics gamma-distribution
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
asked Oct 16 '17 at 6:13
kriskross
142
 |Â
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up vote
0
down vote
favorite
https://i.gyazo.com/9ab33847b7a2f8f7378f12e1ab5960b9.png
A continuous random variable $X$ has the $mathsfGamma(alpha, lambda)$ distribution with probability density
$$f_X(x) = fraclambda^alphaGamma(alpha)x^alpha - 1e^-lambda x,
text for x > 0.$$
Prove from first principles that $E(X^3) = fracalpha(alpha + 1)(alpha + 2)lambda^3.$
I don't know what it means when it says from first principles. Any help? Thanks.
statistics gamma-distribution
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
asked Oct 16 '17 at 6:13
kriskross
142
1
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
1
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
https://i.gyazo.com/9ab33847b7a2f8f7378f12e1ab5960b9.png
A continuous random variable $X$ has the $mathsfGamma(alpha, lambda)$ distribution with probability density
$$f_X(x) = fraclambda^alphaGamma(alpha)x^alpha - 1e^-lambda x,
text for x > 0.$$
Prove from first principles that $E(X^3) = fracalpha(alpha + 1)(alpha + 2)lambda^3.$
I don't know what it means when it says from first principles. Any help? Thanks.
statistics gamma-distribution
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
asked Oct 16 '17 at 6:13
kriskross
142
https://i.gyazo.com/9ab33847b7a2f8f7378f12e1ab5960b9.png
A continuous random variable $X$ has the $mathsfGamma(alpha, lambda)$ distribution with probability density
$$f_X(x) = fraclambda^alphaGamma(alpha)x^alpha - 1e^-lambda x,
text for x > 0.$$
Prove from first principles that $E(X^3) = fracalpha(alpha + 1)(alpha + 2)lambda^3.$
I don't know what it means when it says from first principles. Any help? Thanks.
statistics gamma-distribution
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
asked Oct 16 '17 at 6:13
kriskross
142
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
edited Oct 16 '17 at 6:49
Jean Marie
27.8k41847
27.8k41847
asked Oct 16 '17 at 6:13
kriskross
142
asked Oct 16 '17 at 6:13
kriskross
142
asked Oct 16 '17 at 6:13
kriskross
142
142
1
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
1
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51
 |Â
show 1 more comment
1
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
1
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51
1
1
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
1
1
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51
 |Â
show 1 more comment
1 Answer
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Given that $$f(x)=fraclambda^alphaGamma(alpha,lambda)x^alpha - 1e^-lambda x,
text for x > 0$$
and
$$Gamma(alpha,lambda)=lambda^alpha int_0^infty x^alpha-1 e^-lambda x,dx = lambda^alpha left( left[ fracx^alpha e^-lambda xalpha right]_0^infty + int_0^infty lambda fracx^alpha e^-lambda xalpha,dx right) = frac1alpha Gamma(alpha+1,lambda).$$
We see that $$E(X^3)=int_0^infty x^3 f(x),dx = fracint_0^infty lambda^alpha x^alpha+2 e^-lambda x,dxGamma(alpha,lambda).$$
Note that $int_0^infty lambda^alpha+3x^alpha+2 e^-lambda x,dx = Gamma(alpha+3,lambda)$ so our numerator is $fracGamma(alpha+3,lambda)lambda^3$ and
$$E(X^3) = fracGamma(alpha+3,lambda)lambda^3 Gamma(alpha,lambda) = fracalpha(alpha+1)(alpha+2)lambda^3.$$
QED.
answered Aug 6 at 22:58
Glassjawed
540414
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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up vote
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down vote
Given that $$f(x)=fraclambda^alphaGamma(alpha,lambda)x^alpha - 1e^-lambda x,
text for x > 0$$
and
$$Gamma(alpha,lambda)=lambda^alpha int_0^infty x^alpha-1 e^-lambda x,dx = lambda^alpha left( left[ fracx^alpha e^-lambda xalpha right]_0^infty + int_0^infty lambda fracx^alpha e^-lambda xalpha,dx right) = frac1alpha Gamma(alpha+1,lambda).$$
We see that $$E(X^3)=int_0^infty x^3 f(x),dx = fracint_0^infty lambda^alpha x^alpha+2 e^-lambda x,dxGamma(alpha,lambda).$$
Note that $int_0^infty lambda^alpha+3x^alpha+2 e^-lambda x,dx = Gamma(alpha+3,lambda)$ so our numerator is $fracGamma(alpha+3,lambda)lambda^3$ and
$$E(X^3) = fracGamma(alpha+3,lambda)lambda^3 Gamma(alpha,lambda) = fracalpha(alpha+1)(alpha+2)lambda^3.$$
QED.
answered Aug 6 at 22:58
Glassjawed
540414
add a comment |Â
up vote
0
down vote
Given that $$f(x)=fraclambda^alphaGamma(alpha,lambda)x^alpha - 1e^-lambda x,
text for x > 0$$
and
$$Gamma(alpha,lambda)=lambda^alpha int_0^infty x^alpha-1 e^-lambda x,dx = lambda^alpha left( left[ fracx^alpha e^-lambda xalpha right]_0^infty + int_0^infty lambda fracx^alpha e^-lambda xalpha,dx right) = frac1alpha Gamma(alpha+1,lambda).$$
We see that $$E(X^3)=int_0^infty x^3 f(x),dx = fracint_0^infty lambda^alpha x^alpha+2 e^-lambda x,dxGamma(alpha,lambda).$$
Note that $int_0^infty lambda^alpha+3x^alpha+2 e^-lambda x,dx = Gamma(alpha+3,lambda)$ so our numerator is $fracGamma(alpha+3,lambda)lambda^3$ and
$$E(X^3) = fracGamma(alpha+3,lambda)lambda^3 Gamma(alpha,lambda) = fracalpha(alpha+1)(alpha+2)lambda^3.$$
QED.
answered Aug 6 at 22:58
Glassjawed
540414
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given that $$f(x)=fraclambda^alphaGamma(alpha,lambda)x^alpha - 1e^-lambda x,
text for x > 0$$
and
$$Gamma(alpha,lambda)=lambda^alpha int_0^infty x^alpha-1 e^-lambda x,dx = lambda^alpha left( left[ fracx^alpha e^-lambda xalpha right]_0^infty + int_0^infty lambda fracx^alpha e^-lambda xalpha,dx right) = frac1alpha Gamma(alpha+1,lambda).$$
We see that $$E(X^3)=int_0^infty x^3 f(x),dx = fracint_0^infty lambda^alpha x^alpha+2 e^-lambda x,dxGamma(alpha,lambda).$$
Note that $int_0^infty lambda^alpha+3x^alpha+2 e^-lambda x,dx = Gamma(alpha+3,lambda)$ so our numerator is $fracGamma(alpha+3,lambda)lambda^3$ and
$$E(X^3) = fracGamma(alpha+3,lambda)lambda^3 Gamma(alpha,lambda) = fracalpha(alpha+1)(alpha+2)lambda^3.$$
QED.
answered Aug 6 at 22:58
Glassjawed
540414
Given that $$f(x)=fraclambda^alphaGamma(alpha,lambda)x^alpha - 1e^-lambda x,
text for x > 0$$
and
$$Gamma(alpha,lambda)=lambda^alpha int_0^infty x^alpha-1 e^-lambda x,dx = lambda^alpha left( left[ fracx^alpha e^-lambda xalpha right]_0^infty + int_0^infty lambda fracx^alpha e^-lambda xalpha,dx right) = frac1alpha Gamma(alpha+1,lambda).$$
We see that $$E(X^3)=int_0^infty x^3 f(x),dx = fracint_0^infty lambda^alpha x^alpha+2 e^-lambda x,dxGamma(alpha,lambda).$$
Note that $int_0^infty lambda^alpha+3x^alpha+2 e^-lambda x,dx = Gamma(alpha+3,lambda)$ so our numerator is $fracGamma(alpha+3,lambda)lambda^3$ and
$$E(X^3) = fracGamma(alpha+3,lambda)lambda^3 Gamma(alpha,lambda) = fracalpha(alpha+1)(alpha+2)lambda^3.$$
QED.
answered Aug 6 at 22:58
Glassjawed
540414
edited Aug 6 at 23:36
answered Aug 6 at 22:58
Glassjawed
540414
answered Aug 6 at 22:58
Glassjawed
540414
answered Aug 6 at 22:58
Glassjawed
540414
540414
add a comment |Â
add a comment |Â
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1
I edited the problem in the image directly into the problem. Please check that it is accurate. \ I suppose you are asked to express $E(X^3)$ as an integral involving the density function and do the integration.
– BruceET
Oct 16 '17 at 6:44
1
I have edited your title in order it reflects the content of the question
– Jean Marie
Oct 16 '17 at 6:49
Note that $int x^3 f_alpha (x) dx$ is very similar to $int f_alpha +3(x) dx $ if you adjust the constants at the front of $f$.
– user121049
Oct 16 '17 at 6:58
@user121049 So doing the integral in this situation means to do it from first principles? It was just that wording that was throwing me off. I did do the integral and found $$intlimits_0^infty(lambda^alpha+3x^alpha+2e^-lambdax/Gamma(alpha+3)) dx = 1$$ and so I ended up with $$E[X^3] = Gamma(alpha+3)/lambda^3Gamma(alpha)$$ Is this valid and means I worked it out from first principle?
– kriskross
Oct 16 '17 at 7:47
"First principles" is a bit open ended. How far back is one expected to go? I think your answer is good. I guess they just didn't want you to look the answer up in a book.
– user121049
Oct 16 '17 at 7:51