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Integral boudaries when calculating expectations
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Random variables $X_1,...X_n$ are independent and identically distributed with $X_1$ having the probability density function:
$$ f(x;theta) = 3theta^3 frac1x^4 space for space x>0 $$
Where $theta>0$ Is an unknown parameter.
Then :
$$E[X_1]= 3theta^3 int_theta^infty x x^-4 dx $$
Why is the lower boundry of the integral $theta$ and not $0$?
integration definite-integrals expectation
asked Jul 30 at 9:30
user1607
608
add a comment |Â
up vote
0
down vote
favorite
Random variables $X_1,...X_n$ are independent and identically distributed with $X_1$ having the probability density function:
$$ f(x;theta) = 3theta^3 frac1x^4 space for space x>0 $$
Where $theta>0$ Is an unknown parameter.
Then :
$$E[X_1]= 3theta^3 int_theta^infty x x^-4 dx $$
Why is the lower boundry of the integral $theta$ and not $0$?
integration definite-integrals expectation
asked Jul 30 at 9:30
user1607
608
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Random variables $X_1,...X_n$ are independent and identically distributed with $X_1$ having the probability density function:
$$ f(x;theta) = 3theta^3 frac1x^4 space for space x>0 $$
Where $theta>0$ Is an unknown parameter.
Then :
$$E[X_1]= 3theta^3 int_theta^infty x x^-4 dx $$
Why is the lower boundry of the integral $theta$ and not $0$?
integration definite-integrals expectation
asked Jul 30 at 9:30
user1607
608
Random variables $X_1,...X_n$ are independent and identically distributed with $X_1$ having the probability density function:
$$ f(x;theta) = 3theta^3 frac1x^4 space for space x>0 $$
Where $theta>0$ Is an unknown parameter.
Then :
$$E[X_1]= 3theta^3 int_theta^infty x x^-4 dx $$
Why is the lower boundry of the integral $theta$ and not $0$?
integration definite-integrals expectation
asked Jul 30 at 9:30
user1607
608
asked Jul 30 at 9:30
user1607
608
asked Jul 30 at 9:30
user1607
608
asked Jul 30 at 9:30
user1607
608
608
add a comment |Â
add a comment |Â
1 Answer
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The definition of $f(x)$ should be for $x>theta$ because $int_theta^inftyf(x)dx=1$.
answered Jul 30 at 9:35
Empy2
31.7k12059
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The definition of $f(x)$ should be for $x>theta$ because $int_theta^inftyf(x)dx=1$.
answered Jul 30 at 9:35
Empy2
31.7k12059
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
add a comment |Â
up vote
0
down vote
The definition of $f(x)$ should be for $x>theta$ because $int_theta^inftyf(x)dx=1$.
answered Jul 30 at 9:35
Empy2
31.7k12059
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The definition of $f(x)$ should be for $x>theta$ because $int_theta^inftyf(x)dx=1$.
answered Jul 30 at 9:35
Empy2
31.7k12059
The definition of $f(x)$ should be for $x>theta$ because $int_theta^inftyf(x)dx=1$.
answered Jul 30 at 9:35
Empy2
31.7k12059
answered Jul 30 at 9:35
Empy2
31.7k12059
answered Jul 30 at 9:35
Empy2
31.7k12059
answered Jul 30 at 9:35
Empy2
31.7k12059
31.7k12059
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
add a comment |Â
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
yes, I was going to say the same.
– Francisco
Jul 30 at 9:40
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
Must be a typo in the exercise, nonethless we know that $theta$ > 0, therefore could 0 be used for the lower boundry insted of the $theta$ ?
– user1607
Jul 30 at 9:58
add a comment |Â
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