Trying to Understand $E[X^2]$ for Gamma Distribution

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I am trying to understand the following for the gamma distribution:

$$E[X^2] = frac alpha(alpha+1)lambda^2$$

I've been looking at the reasoning for $E[X]$ to make sense of what could be happening, for instance:

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 = fracalphalambda$$

I can't make sense of what is happening to the second line where the integral drops out to become $$int_0^inftyx^alpha-1+1e^-lambda xdx =fracGamma(alpha+1)lambda^alpha+1$$
I figure understanding this point would help with understanding the reasoning behind $E[X^2]$.

Thank you!

probability expectation gamma-function gamma-distribution

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asked Jul 19 at 13:22

M. Damon

132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  suggest you check the definition of a gamma function.
  – John Polcari
  Jul 19 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also the laplace transform of the function $f(t)=t^alpha$
  – Isham
  Jul 19 at 13:36
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I am trying to understand the following for the gamma distribution:

$$E[X^2] = frac alpha(alpha+1)lambda^2$$

I've been looking at the reasoning for $E[X]$ to make sense of what could be happening, for instance:

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 = fracalphalambda$$

I can't make sense of what is happening to the second line where the integral drops out to become $$int_0^inftyx^alpha-1+1e^-lambda xdx =fracGamma(alpha+1)lambda^alpha+1$$
I figure understanding this point would help with understanding the reasoning behind $E[X^2]$.

Thank you!

probability expectation gamma-function gamma-distribution

share|cite|improve this question

asked Jul 19 at 13:22

M. Damon

132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  suggest you check the definition of a gamma function.
  – John Polcari
  Jul 19 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also the laplace transform of the function $f(t)=t^alpha$
  – Isham
  Jul 19 at 13:36
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am trying to understand the following for the gamma distribution:

$$E[X^2] = frac alpha(alpha+1)lambda^2$$

I've been looking at the reasoning for $E[X]$ to make sense of what could be happening, for instance:

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 = fracalphalambda$$

I can't make sense of what is happening to the second line where the integral drops out to become $$int_0^inftyx^alpha-1+1e^-lambda xdx =fracGamma(alpha+1)lambda^alpha+1$$
I figure understanding this point would help with understanding the reasoning behind $E[X^2]$.

Thank you!

probability expectation gamma-function gamma-distribution

share|cite|improve this question

asked Jul 19 at 13:22

M. Damon

132

I am trying to understand the following for the gamma distribution:

$$E[X^2] = frac alpha(alpha+1)lambda^2$$

I've been looking at the reasoning for $E[X]$ to make sense of what could be happening, for instance:

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 = fracalphalambda$$

I can't make sense of what is happening to the second line where the integral drops out to become $$int_0^inftyx^alpha-1+1e^-lambda xdx =fracGamma(alpha+1)lambda^alpha+1$$
I figure understanding this point would help with understanding the reasoning behind $E[X^2]$.

Thank you!

probability expectation gamma-function gamma-distribution

share|cite|improve this question

asked Jul 19 at 13:22

M. Damon

132

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 13:22

M. Damon

132

asked Jul 19 at 13:22

M. Damon

132

asked Jul 19 at 13:22

M. Damon

132

132

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  suggest you check the definition of a gamma function.
  – John Polcari
  Jul 19 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also the laplace transform of the function $f(t)=t^alpha$
  – Isham
  Jul 19 at 13:36
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  suggest you check the definition of a gamma function.
  – John Polcari
  Jul 19 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also the laplace transform of the function $f(t)=t^alpha$
  – Isham
  Jul 19 at 13:36
  
  

  
  
  
  

2

2

suggest you check the definition of a gamma function.
– John Polcari
Jul 19 at 13:25

suggest you check the definition of a gamma function.
– John Polcari
Jul 19 at 13:25

It's also the laplace transform of the function $f(t)=t^alpha$
– Isham
Jul 19 at 13:36

It's also the laplace transform of the function $f(t)=t^alpha$
– Isham
Jul 19 at 13:36

add a comment | 

2 Answers
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HINT:

The gamma function is defined as $$Gamma(t)=int_0^infty x^t-1e^-x,dx$$ so let $u=lambda x$ and $t=alpha+1$.

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

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up vote
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$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx = fraclambda^alphaGamma(alpha) fraclambda^a+1Gamma(a+1)lambda^a+1Gamma(a+1)int_0^inftyx^alpha-1+1e^-lambda xdx = $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 int_0^infty frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfracdxx = fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 times 1 = fracalambda$$
because $f(x) = frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfrac1x$ is a PDF of $X sim Gamma(a+1, lambda)$

share|cite|improve this answer

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

HINT:

The gamma function is defined as $$Gamma(t)=int_0^infty x^t-1e^-x,dx$$ so let $u=lambda x$ and $t=alpha+1$.

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

add a comment | 

up vote
2
down vote

HINT:

The gamma function is defined as $$Gamma(t)=int_0^infty x^t-1e^-x,dx$$ so let $u=lambda x$ and $t=alpha+1$.

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

add a comment | 

up vote
2
down vote

up vote
2
down vote

HINT:

The gamma function is defined as $$Gamma(t)=int_0^infty x^t-1e^-x,dx$$ so let $u=lambda x$ and $t=alpha+1$.

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

HINT:

The gamma function is defined as $$Gamma(t)=int_0^infty x^t-1e^-x,dx$$ so let $u=lambda x$ and $t=alpha+1$.

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

answered Jul 19 at 13:27

TheSimpliFire

9,67461951

9,67461951

add a comment | 

add a comment | 

up vote
2
down vote

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx = fraclambda^alphaGamma(alpha) fraclambda^a+1Gamma(a+1)lambda^a+1Gamma(a+1)int_0^inftyx^alpha-1+1e^-lambda xdx = $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 int_0^infty frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfracdxx = fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 times 1 = fracalambda$$
because $f(x) = frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfrac1x$ is a PDF of $X sim Gamma(a+1, lambda)$

share|cite|improve this answer

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

add a comment | 

up vote
2
down vote

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx = fraclambda^alphaGamma(alpha) fraclambda^a+1Gamma(a+1)lambda^a+1Gamma(a+1)int_0^inftyx^alpha-1+1e^-lambda xdx = $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 int_0^infty frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfracdxx = fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 times 1 = fracalambda$$
because $f(x) = frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfrac1x$ is a PDF of $X sim Gamma(a+1, lambda)$

share|cite|improve this answer

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

add a comment | 

up vote
2
down vote

up vote
2
down vote

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx = fraclambda^alphaGamma(alpha) fraclambda^a+1Gamma(a+1)lambda^a+1Gamma(a+1)int_0^inftyx^alpha-1+1e^-lambda xdx = $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 int_0^infty frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfracdxx = fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 times 1 = fracalambda$$
because $f(x) = frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfrac1x$ is a PDF of $X sim Gamma(a+1, lambda)$

share|cite|improve this answer

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

$$E[X] = int_0^inftyfraclambda^alphaGamma(alpha)xcdot x^alpha-1e^-lambda xdx$$
$$=fraclambda^alphaGamma(alpha)int_0^inftyx^alpha-1+1e^-lambda xdx = fraclambda^alphaGamma(alpha) fraclambda^a+1Gamma(a+1)lambda^a+1Gamma(a+1)int_0^inftyx^alpha-1+1e^-lambda xdx = $$
$$=fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 int_0^infty frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfracdxx = fraclambda^alphaGamma(alpha)cdot fracGamma(alpha+1)lambda^alpha+1 times 1 = fracalambda$$
because $f(x) = frac1Gamma(a+1) (lambda x) ^a+1 e^-lambda xfrac1x$ is a PDF of $X sim Gamma(a+1, lambda)$

share|cite|improve this answer

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 13:42

edited Jul 19 at 13:42

edited Jul 19 at 13:42

answered Jul 19 at 13:40

D F

1,0551218

answered Jul 19 at 13:40

D F

1,0551218

answered Jul 19 at 13:40

D F

1,0551218

1,0551218

add a comment | 

add a comment | 

 

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