A Problem finding a density Function
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I made up the following problem. Is my answer correct? Any comments about my style?
Thanks,
Bob
Problem:
Find the family of densities functions which are parabolas of the form
$y = Ax^2 + Bx + C$, have their vertex on the y axis, intersect the x-axis at the points $(pm c, 0)$ where $c$ is a positive real number and $A < 0$. In addition, the density function should be $0$ for $x < -c$ and $x > c$. Also find $E(x^2)$,
$E(x^3)$ and the standard deviation of $x$.
Answer:
The first step in the process is to find the density function. Since we have
$A < 0$, the vertex is going to be a maximum, not a minimum.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
A(-c)^2 + B(-c) + C &=& 0 \
Ac^2 - Bc + C &=& 0 \
C &=& -Ac^2 \
endeqnarray*
Now since we are looking for a density function, the area under the curve must
be $1$.
begineqnarray*
int_-c^c Ax^2 + Bx + C , dx &=& 1 \
fracAx^33 + fracBx^22 + Cx Big|_-c^c &=& 1 \
fracAc^33 + fracBc^22 + Cc - ( frac-Ac^33 + fracBc^22 - Cc )
&=& 1 \
frac2Ac^33 + 2Cc &=& 1 \
frac2Ac^33 + 2(-Ac^2)c &=& 1 \
endeqnarray*
begineqnarray*
fracAc^33 - Ac^3 &=& 1 \
-frac2Ac^33 &=& 1 \
A &=& frac-32c^3 \
C &=& -Ac^2 = Big( frac32c^2 Big) (c^2 ) \
C = frac32c
endeqnarray*
Now we need to solve for $B$.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
Big( frac-32c^3Big)c^2 + Bc + frac32c &=& 0 \
frac-32c + Bc + frac32c &=& 0 \
Bc &=& 0 \
B &=& 0 \
endeqnarray*
Hence our density function is:
begineqnarray*
y &=& frac-32c^3x^2 + frac32c ,,, text for -c <= x <= c \
E(x^2) &=& int_-c^c x^2 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^2) &=& int_-c^c frac-3x^42c^3 + frac3x^22c ,, dx =
frac-3x^510c^3 + frac3x^36c Big|_-c^c \
E(x^2) &=&
frac-3c^510c^3 + frac3c^36c - ( frac3c^510c^3 + frac-3c^36c ) \
E(x^2) &=&
frac-3c^510c^3 + fracc^32c - frac3c^510c^3 + fracc^32c \
E(x^2) &=&
frac-3c^210 + fracc^32c - frac3c^210 + fracc^32c \
E(x^2) &=& fracc^3c - frac3c^210 - frac3c^210 \
E(x^2) &=& c^2 - frac6c^210 \
E(x^2) &=& frac2c^25 \
endeqnarray*
begineqnarray*
E(x^3) &=& int_-c^c x^3 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^3) &=& int_-c^c frac-32c^3x^5 + frac3x^32c ,, dx \
E(x^3) &=& frac-3x^612c^3 + frac3x^48c Big|_-c^c \
E(x^3) &=&
frac-3c^612c^3 + frac3c^48c -
Big( frac-3c^612c^3 + frac3c^48c Big) \
E(x^3) &=& 0 \
sigma^2 &=& E(x^2) - u^2 = E(x^2 ) \
sigma^2 &=& frac2c^25 \
sigma &=& fracsqrt2csqrt5 \
endeqnarray*
probability probability-distributions
asked Aug 26 at 21:20
Bob
778411
add a comment |Â
up vote
3
down vote
favorite
I made up the following problem. Is my answer correct? Any comments about my style?
Thanks,
Bob
Problem:
Find the family of densities functions which are parabolas of the form
$y = Ax^2 + Bx + C$, have their vertex on the y axis, intersect the x-axis at the points $(pm c, 0)$ where $c$ is a positive real number and $A < 0$. In addition, the density function should be $0$ for $x < -c$ and $x > c$. Also find $E(x^2)$,
$E(x^3)$ and the standard deviation of $x$.
Answer:
The first step in the process is to find the density function. Since we have
$A < 0$, the vertex is going to be a maximum, not a minimum.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
A(-c)^2 + B(-c) + C &=& 0 \
Ac^2 - Bc + C &=& 0 \
C &=& -Ac^2 \
endeqnarray*
Now since we are looking for a density function, the area under the curve must
be $1$.
begineqnarray*
int_-c^c Ax^2 + Bx + C , dx &=& 1 \
fracAx^33 + fracBx^22 + Cx Big|_-c^c &=& 1 \
fracAc^33 + fracBc^22 + Cc - ( frac-Ac^33 + fracBc^22 - Cc )
&=& 1 \
frac2Ac^33 + 2Cc &=& 1 \
frac2Ac^33 + 2(-Ac^2)c &=& 1 \
endeqnarray*
begineqnarray*
fracAc^33 - Ac^3 &=& 1 \
-frac2Ac^33 &=& 1 \
A &=& frac-32c^3 \
C &=& -Ac^2 = Big( frac32c^2 Big) (c^2 ) \
C = frac32c
endeqnarray*
Now we need to solve for $B$.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
Big( frac-32c^3Big)c^2 + Bc + frac32c &=& 0 \
frac-32c + Bc + frac32c &=& 0 \
Bc &=& 0 \
B &=& 0 \
endeqnarray*
Hence our density function is:
begineqnarray*
y &=& frac-32c^3x^2 + frac32c ,,, text for -c <= x <= c \
E(x^2) &=& int_-c^c x^2 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^2) &=& int_-c^c frac-3x^42c^3 + frac3x^22c ,, dx =
frac-3x^510c^3 + frac3x^36c Big|_-c^c \
E(x^2) &=&
frac-3c^510c^3 + frac3c^36c - ( frac3c^510c^3 + frac-3c^36c ) \
E(x^2) &=&
frac-3c^510c^3 + fracc^32c - frac3c^510c^3 + fracc^32c \
E(x^2) &=&
frac-3c^210 + fracc^32c - frac3c^210 + fracc^32c \
E(x^2) &=& fracc^3c - frac3c^210 - frac3c^210 \
E(x^2) &=& c^2 - frac6c^210 \
E(x^2) &=& frac2c^25 \
endeqnarray*
begineqnarray*
E(x^3) &=& int_-c^c x^3 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^3) &=& int_-c^c frac-32c^3x^5 + frac3x^32c ,, dx \
E(x^3) &=& frac-3x^612c^3 + frac3x^48c Big|_-c^c \
E(x^3) &=&
frac-3c^612c^3 + frac3c^48c -
Big( frac-3c^612c^3 + frac3c^48c Big) \
E(x^3) &=& 0 \
sigma^2 &=& E(x^2) - u^2 = E(x^2 ) \
sigma^2 &=& frac2c^25 \
sigma &=& fracsqrt2csqrt5 \
endeqnarray*
probability probability-distributions
asked Aug 26 at 21:20
Bob
778411
1
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I made up the following problem. Is my answer correct? Any comments about my style?
Thanks,
Bob
Problem:
Find the family of densities functions which are parabolas of the form
$y = Ax^2 + Bx + C$, have their vertex on the y axis, intersect the x-axis at the points $(pm c, 0)$ where $c$ is a positive real number and $A < 0$. In addition, the density function should be $0$ for $x < -c$ and $x > c$. Also find $E(x^2)$,
$E(x^3)$ and the standard deviation of $x$.
Answer:
The first step in the process is to find the density function. Since we have
$A < 0$, the vertex is going to be a maximum, not a minimum.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
A(-c)^2 + B(-c) + C &=& 0 \
Ac^2 - Bc + C &=& 0 \
C &=& -Ac^2 \
endeqnarray*
Now since we are looking for a density function, the area under the curve must
be $1$.
begineqnarray*
int_-c^c Ax^2 + Bx + C , dx &=& 1 \
fracAx^33 + fracBx^22 + Cx Big|_-c^c &=& 1 \
fracAc^33 + fracBc^22 + Cc - ( frac-Ac^33 + fracBc^22 - Cc )
&=& 1 \
frac2Ac^33 + 2Cc &=& 1 \
frac2Ac^33 + 2(-Ac^2)c &=& 1 \
endeqnarray*
begineqnarray*
fracAc^33 - Ac^3 &=& 1 \
-frac2Ac^33 &=& 1 \
A &=& frac-32c^3 \
C &=& -Ac^2 = Big( frac32c^2 Big) (c^2 ) \
C = frac32c
endeqnarray*
Now we need to solve for $B$.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
Big( frac-32c^3Big)c^2 + Bc + frac32c &=& 0 \
frac-32c + Bc + frac32c &=& 0 \
Bc &=& 0 \
B &=& 0 \
endeqnarray*
Hence our density function is:
begineqnarray*
y &=& frac-32c^3x^2 + frac32c ,,, text for -c <= x <= c \
E(x^2) &=& int_-c^c x^2 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^2) &=& int_-c^c frac-3x^42c^3 + frac3x^22c ,, dx =
frac-3x^510c^3 + frac3x^36c Big|_-c^c \
E(x^2) &=&
frac-3c^510c^3 + frac3c^36c - ( frac3c^510c^3 + frac-3c^36c ) \
E(x^2) &=&
frac-3c^510c^3 + fracc^32c - frac3c^510c^3 + fracc^32c \
E(x^2) &=&
frac-3c^210 + fracc^32c - frac3c^210 + fracc^32c \
E(x^2) &=& fracc^3c - frac3c^210 - frac3c^210 \
E(x^2) &=& c^2 - frac6c^210 \
E(x^2) &=& frac2c^25 \
endeqnarray*
begineqnarray*
E(x^3) &=& int_-c^c x^3 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^3) &=& int_-c^c frac-32c^3x^5 + frac3x^32c ,, dx \
E(x^3) &=& frac-3x^612c^3 + frac3x^48c Big|_-c^c \
E(x^3) &=&
frac-3c^612c^3 + frac3c^48c -
Big( frac-3c^612c^3 + frac3c^48c Big) \
E(x^3) &=& 0 \
sigma^2 &=& E(x^2) - u^2 = E(x^2 ) \
sigma^2 &=& frac2c^25 \
sigma &=& fracsqrt2csqrt5 \
endeqnarray*
probability probability-distributions
asked Aug 26 at 21:20
Bob
778411
I made up the following problem. Is my answer correct? Any comments about my style?
Thanks,
Bob
Problem:
Find the family of densities functions which are parabolas of the form
$y = Ax^2 + Bx + C$, have their vertex on the y axis, intersect the x-axis at the points $(pm c, 0)$ where $c$ is a positive real number and $A < 0$. In addition, the density function should be $0$ for $x < -c$ and $x > c$. Also find $E(x^2)$,
$E(x^3)$ and the standard deviation of $x$.
Answer:
The first step in the process is to find the density function. Since we have
$A < 0$, the vertex is going to be a maximum, not a minimum.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
A(-c)^2 + B(-c) + C &=& 0 \
Ac^2 - Bc + C &=& 0 \
C &=& -Ac^2 \
endeqnarray*
Now since we are looking for a density function, the area under the curve must
be $1$.
begineqnarray*
int_-c^c Ax^2 + Bx + C , dx &=& 1 \
fracAx^33 + fracBx^22 + Cx Big|_-c^c &=& 1 \
fracAc^33 + fracBc^22 + Cc - ( frac-Ac^33 + fracBc^22 - Cc )
&=& 1 \
frac2Ac^33 + 2Cc &=& 1 \
frac2Ac^33 + 2(-Ac^2)c &=& 1 \
endeqnarray*
begineqnarray*
fracAc^33 - Ac^3 &=& 1 \
-frac2Ac^33 &=& 1 \
A &=& frac-32c^3 \
C &=& -Ac^2 = Big( frac32c^2 Big) (c^2 ) \
C = frac32c
endeqnarray*
Now we need to solve for $B$.
begineqnarray*
Ac^2 + Bc + C &=& 0 \
Big( frac-32c^3Big)c^2 + Bc + frac32c &=& 0 \
frac-32c + Bc + frac32c &=& 0 \
Bc &=& 0 \
B &=& 0 \
endeqnarray*
Hence our density function is:
begineqnarray*
y &=& frac-32c^3x^2 + frac32c ,,, text for -c <= x <= c \
E(x^2) &=& int_-c^c x^2 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^2) &=& int_-c^c frac-3x^42c^3 + frac3x^22c ,, dx =
frac-3x^510c^3 + frac3x^36c Big|_-c^c \
E(x^2) &=&
frac-3c^510c^3 + frac3c^36c - ( frac3c^510c^3 + frac-3c^36c ) \
E(x^2) &=&
frac-3c^510c^3 + fracc^32c - frac3c^510c^3 + fracc^32c \
E(x^2) &=&
frac-3c^210 + fracc^32c - frac3c^210 + fracc^32c \
E(x^2) &=& fracc^3c - frac3c^210 - frac3c^210 \
E(x^2) &=& c^2 - frac6c^210 \
E(x^2) &=& frac2c^25 \
endeqnarray*
begineqnarray*
E(x^3) &=& int_-c^c x^3 Big( frac-32c^3x^2 + frac32c Big) ,, dx \
E(x^3) &=& int_-c^c frac-32c^3x^5 + frac3x^32c ,, dx \
E(x^3) &=& frac-3x^612c^3 + frac3x^48c Big|_-c^c \
E(x^3) &=&
frac-3c^612c^3 + frac3c^48c -
Big( frac-3c^612c^3 + frac3c^48c Big) \
E(x^3) &=& 0 \
sigma^2 &=& E(x^2) - u^2 = E(x^2 ) \
sigma^2 &=& frac2c^25 \
sigma &=& fracsqrt2csqrt5 \
endeqnarray*
probability probability-distributions
asked Aug 26 at 21:20
Bob
778411
edited Aug 27 at 19:11
asked Aug 26 at 21:20
Bob
778411
asked Aug 26 at 21:20
Bob
778411
asked Aug 26 at 21:20
Bob
778411
778411
1
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40
add a comment |Â
1
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40
1
1
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40
add a comment |Â
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1
Since the parabola is symmetric (x intercepts at$pm c$), B=0. Also all odd moments $E(x^n)=0$ for odd n, also by symmetry.
– herb steinberg
Aug 26 at 21:58
We know that $pm c$ are zeroes, therefore $$f_X(x) = alpha (c^2 - x^2), \ alpha = left( 2 int_0^c (c^2 - x^2) dx right)^-1, \ operatorname E[X^2] = 2 alpha int_0^c x^2 (c^2 - x^2) dx.$$ You have a typo in the computation of $alpha$, $A c^3 / 3 - A c^3 = 1/2$, not $1$.
– Maxim
Aug 27 at 20:40