Let $Y_n = fracsum_i=1^n X_i(X_i+1)n$ . Find a number $c$ such that $Y_n to c$
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Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random variables, each
having a $Uniform[0, 0.5]$ distribution. Let $Y_n = fracsum_i=1^n X_i(X_i+1)n$
. Find a number $c$ such that
$Y_n to c$ ($p$ above the arrow)
Attempt:
$Y_n = fracsum_i=0^n X_i(X_i+1)n = fracsum_i=1^n X_i^2n + fracsum_i=1^n X_in$
Not sure how to go about this. I believe I need to find their expectation but not sure how.
probability expectation
asked Aug 11 at 0:35
Bas bas
39611
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0
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Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random variables, each
having a $Uniform[0, 0.5]$ distribution. Let $Y_n = fracsum_i=1^n X_i(X_i+1)n$
. Find a number $c$ such that
$Y_n to c$ ($p$ above the arrow)
Attempt:
$Y_n = fracsum_i=0^n X_i(X_i+1)n = fracsum_i=1^n X_i^2n + fracsum_i=1^n X_in$
Not sure how to go about this. I believe I need to find their expectation but not sure how.
probability expectation
asked Aug 11 at 0:35
Bas bas
39611
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random variables, each
having a $Uniform[0, 0.5]$ distribution. Let $Y_n = fracsum_i=1^n X_i(X_i+1)n$
. Find a number $c$ such that
$Y_n to c$ ($p$ above the arrow)
Attempt:
$Y_n = fracsum_i=0^n X_i(X_i+1)n = fracsum_i=1^n X_i^2n + fracsum_i=1^n X_in$
Not sure how to go about this. I believe I need to find their expectation but not sure how.
probability expectation
asked Aug 11 at 0:35
Bas bas
39611
Let $X_1, X_2, . . . , X_n$ be independent and identically distributed random variables, each
having a $Uniform[0, 0.5]$ distribution. Let $Y_n = fracsum_i=1^n X_i(X_i+1)n$
. Find a number $c$ such that
$Y_n to c$ ($p$ above the arrow)
Attempt:
$Y_n = fracsum_i=0^n X_i(X_i+1)n = fracsum_i=1^n X_i^2n + fracsum_i=1^n X_in$
Not sure how to go about this. I believe I need to find their expectation but not sure how.
probability expectation
asked Aug 11 at 0:35
Bas bas
39611
asked Aug 11 at 0:35
Bas bas
39611
asked Aug 11 at 0:35
Bas bas
39611
asked Aug 11 at 0:35
Bas bas
39611
39611
add a comment |Â
add a comment |Â
2 Answers
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The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 int _0^ 1/2 x , dx =frac 1 4$ and $EX_1^2=2int _0^ 1/2 x ^2, dx =frac 1 12$ so $c=frac 1 4+frac 1 12= frac 1 3$.
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
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By the Law of Large Numbers $$fracsum_i=0^n X_in rightarrow mu$$
where $$mu = E(X) = frac12(0 + 0.5) = frac14$$
Also
$$fracsum_i=0^n X_i^2n rightarrow mu^2 + Var(X)$$
where $$Var(X) = frac112(0.5)^2 = frac148 $$
So
$$fracsum_i=0^n X_i^2n rightarrow frac116 + frac148$$
Therefore
$$c = frac14 + frac116 + frac148 = frac13$$
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 int _0^ 1/2 x , dx =frac 1 4$ and $EX_1^2=2int _0^ 1/2 x ^2, dx =frac 1 12$ so $c=frac 1 4+frac 1 12= frac 1 3$.
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
add a comment |Â
up vote
0
down vote
accepted
The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 int _0^ 1/2 x , dx =frac 1 4$ and $EX_1^2=2int _0^ 1/2 x ^2, dx =frac 1 12$ so $c=frac 1 4+frac 1 12= frac 1 3$.
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 int _0^ 1/2 x , dx =frac 1 4$ and $EX_1^2=2int _0^ 1/2 x ^2, dx =frac 1 12$ so $c=frac 1 4+frac 1 12= frac 1 3$.
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
The density is $2$ on $(0,0.5)$ and $0$ elsewhere. $EX_1=2 int _0^ 1/2 x , dx =frac 1 4$ and $EX_1^2=2int _0^ 1/2 x ^2, dx =frac 1 12$ so $c=frac 1 4+frac 1 12= frac 1 3$.
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
answered Aug 11 at 0:41
Kavi Rama Murthy
21.8k2932
21.8k2932
add a comment |Â
add a comment |Â
up vote
1
down vote
By the Law of Large Numbers $$fracsum_i=0^n X_in rightarrow mu$$
where $$mu = E(X) = frac12(0 + 0.5) = frac14$$
Also
$$fracsum_i=0^n X_i^2n rightarrow mu^2 + Var(X)$$
where $$Var(X) = frac112(0.5)^2 = frac148 $$
So
$$fracsum_i=0^n X_i^2n rightarrow frac116 + frac148$$
Therefore
$$c = frac14 + frac116 + frac148 = frac13$$
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
add a comment |Â
up vote
1
down vote
By the Law of Large Numbers $$fracsum_i=0^n X_in rightarrow mu$$
where $$mu = E(X) = frac12(0 + 0.5) = frac14$$
Also
$$fracsum_i=0^n X_i^2n rightarrow mu^2 + Var(X)$$
where $$Var(X) = frac112(0.5)^2 = frac148 $$
So
$$fracsum_i=0^n X_i^2n rightarrow frac116 + frac148$$
Therefore
$$c = frac14 + frac116 + frac148 = frac13$$
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By the Law of Large Numbers $$fracsum_i=0^n X_in rightarrow mu$$
where $$mu = E(X) = frac12(0 + 0.5) = frac14$$
Also
$$fracsum_i=0^n X_i^2n rightarrow mu^2 + Var(X)$$
where $$Var(X) = frac112(0.5)^2 = frac148 $$
So
$$fracsum_i=0^n X_i^2n rightarrow frac116 + frac148$$
Therefore
$$c = frac14 + frac116 + frac148 = frac13$$
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
By the Law of Large Numbers $$fracsum_i=0^n X_in rightarrow mu$$
where $$mu = E(X) = frac12(0 + 0.5) = frac14$$
Also
$$fracsum_i=0^n X_i^2n rightarrow mu^2 + Var(X)$$
where $$Var(X) = frac112(0.5)^2 = frac148 $$
So
$$fracsum_i=0^n X_i^2n rightarrow frac116 + frac148$$
Therefore
$$c = frac14 + frac116 + frac148 = frac13$$
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
answered Aug 11 at 0:45
Ahmad Bazzi
3,0601419
3,0601419
add a comment |Â
add a comment |Â
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