Compute $ E[|X^2-16|] $ where $ X sim U $
$begingroup$
$ X sim U (-4,7) $ with $U$ a continuous uniform distribution.
X is continuous variable. Compute $ E[|X^2 - 16|] $
So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$
But I'm not sure if that's the correct way to compute?
probability-theory uniform-distribution
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
$endgroup$
add a comment |
$begingroup$
$ X sim U (-4,7) $ with $U$ a continuous uniform distribution.
X is continuous variable. Compute $ E[|X^2 - 16|] $
So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$
But I'm not sure if that's the correct way to compute?
probability-theory uniform-distribution
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
$endgroup$
add a comment |
$begingroup$
$ X sim U (-4,7) $ with $U$ a continuous uniform distribution.
X is continuous variable. Compute $ E[|X^2 - 16|] $
So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$
But I'm not sure if that's the correct way to compute?
probability-theory uniform-distribution
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
$endgroup$
$ X sim U (-4,7) $ with $U$ a continuous uniform distribution.
X is continuous variable. Compute $ E[|X^2 - 16|] $
So: $$ E[|X^2 - 16|] = begin{cases} E[X^2 - 16] , 4< X \ E[-X^2+16], X < 4 end{cases} \ $$ My attempt is to calculate each case and then sum them. $$E[X^2 - 16] = sum_{i=4}^7 (X_i^2-16) cdot frac{1}{11} = frac{1}{11} cdot sum_{j=1}^4 (j+3)^2-16 = frac{1}{11} cdot sum_{j=1}^4 j^2 +6j - 7\ = -frac{28}{11}cdot (sum_{j=1}^4 j^2 + 6cdotsum_{j=1}^4 j) = - frac{28}{11} ( frac{4 cdot 5 cdot 9}{6} + 6 cdot frac{4 cdot 5}{2} )$$
But I'm not sure if that's the correct way to compute?
probability-theory uniform-distribution
probability-theory uniform-distribution
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
edited Dec 18 '18 at 17:13
quid♦
37.2k95193
37.2k95193
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
asked Dec 18 '18 at 8:23
bm1125bm1125
65116
65116
add a comment |
add a comment |
1 Answer
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$begingroup$
$X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
add a comment |
$begingroup$
$X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
add a comment |
$begingroup$
$X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$X$ is a random variable and not a constant. You cannot split the computation into the cases $4<X$ and $X<4$. (The probabilities of these events will have an effect on the answer). Also, you treating a continuous random variable as a discrete one. The finite sums you are considering are not related to uniform distribution at all. The correct formula for $E|X^{2}-16|$ is $frac 1 {11} int_{-4}^{7} |x^{2}-16|, dx$. Now split the integral into integral form $-4$ to $4$ and $4$ to $7$.
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
edited Dec 18 '18 at 8:40
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 18 '18 at 8:28
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
add a comment |
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
$begingroup$
Thanks a lot for pointing that out for me!
$endgroup$
– bm1125
Dec 18 '18 at 8:59
add a comment |
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