Continuous Random Variables question
$begingroup$
The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$
The function $g(X)$ is defined by $g(x) = x^2 + x$
Find the value of $E[g(X)]$.
Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.
I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.
probability random-variables expected-value
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
$endgroup$
add a comment |
$begingroup$
The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$
The function $g(X)$ is defined by $g(x) = x^2 + x$
Find the value of $E[g(X)]$.
Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.
I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.
probability random-variables expected-value
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
$endgroup$
$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
1
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42
add a comment |
$begingroup$
The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$
The function $g(X)$ is defined by $g(x) = x^2 + x$
Find the value of $E[g(X)]$.
Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.
I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.
probability random-variables expected-value
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
$endgroup$
The continuous random variable $X$ has p.d.f. $$f(x) =begin{cases}
frac{2(7-x)}{25},&text{for $2leq xleq 7$}\ 0,& text{otherwise}
end{cases} $$
The function $g(X)$ is defined by $g(x) = x^2 + x$
Find the value of $E[g(X)]$.
Is the correct answer to this question $37/2$ or $397/6$? I disagree with the answer the textbook has given.
I believe the answer to be $37/2$, this is for 2 main reasons: Firstly, if you compute the integral of $(x^2 + x) (14/25 - 2x/25)$ with limits $2$ and $7$ then you get $37/2$. Secondly, an easier way. $E[X]$ in this case is $11/3$, and $E[X^2]$ is $89/6$ add them together: $89/6 + 11/3 = 37/2$. However, the textbook says the answer is $397/6$, and then follows up the question using this value.
probability random-variables expected-value
probability random-variables expected-value
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
edited Dec 10 '18 at 4:10
André 3000
12.6k22243
12.6k22243
asked Dec 2 '18 at 2:05
user546944
$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
1
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42
add a comment |
$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
1
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42
$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
1
1
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In addition to your calculations, which appear to be correct:
With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$
Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
In addition to your calculations, which appear to be correct:
With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$
Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
$endgroup$
add a comment |
$begingroup$
In addition to your calculations, which appear to be correct:
With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$
Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
$endgroup$
add a comment |
$begingroup$
In addition to your calculations, which appear to be correct:
With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$
Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
$endgroup$
In addition to your calculations, which appear to be correct:
With probability $1$ we have that $2 < x leq 7.$
Within the interval $2 < x leq 7,$
the function $g(x) = x^2 + x$ has its maximum at $x = 7.$
With probability $1,$ therefore, $g(x) leq g(7) = 56.$
This implies that $E[g(x)] leq 56.$
Observing that $frac{397}{6} > 66,$ we know that $E[g(x)] neq frac{397}{6}.$
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
answered Dec 2 '18 at 2:47
David KDavid K
54.4k343118
54.4k343118
add a comment |
add a comment |
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$begingroup$
Can you explain why do you disagree?
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:06
1
$begingroup$
I believe the answer to be 37/2, this is for 2 main reasons: Firstly, if you compute the integral of (x^2 + x) (14/25 - 2x/25) with limits 2 and 7 then you get 37/2 Secondly, an easier way. E(X) in this case is 11/3, and E(X^2) is 89/6 add them together: 89/6 + 11/3 = 37/2 This is why I think the answer is 37/2. However, the textbook says the answer is 397/6, and then follows up the question using this.
$endgroup$
– user546944
Dec 2 '18 at 2:11
$begingroup$
My calculation also yielded $18.5$. The textbook answer may be wrong.
$endgroup$
– Thomas Shelby
Dec 2 '18 at 2:32
$begingroup$
Your definition of $f(x)$ uses $gt$ not =. Why?
$endgroup$
– herb steinberg
Dec 9 '18 at 3:42