$mathbb{E}[X] ,Var[X] $ of quadratic equation
$begingroup$
I am having some problems to find the solution required in the next exercise:
$x^2-ax+b=0$ has real and not equal solutions and that b is a positive constant that such that $b sim U(i,j)$ for certain i,j.
Find the $mathbb{E}[X], ,Var[X]$ of the solutions.
What I have so far:
From the Quadratic Formula, it is known that:
And since these are real solutions,
$implies a^2>4b$
$implies frac{a^2}{4}>b$
¿how do I use $b sim U(i,j)$?
probability probability-theory statistics probability-distributions normal-distribution
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
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add a comment |
$begingroup$
I am having some problems to find the solution required in the next exercise:
$x^2-ax+b=0$ has real and not equal solutions and that b is a positive constant that such that $b sim U(i,j)$ for certain i,j.
Find the $mathbb{E}[X], ,Var[X]$ of the solutions.
What I have so far:
From the Quadratic Formula, it is known that:
And since these are real solutions,
$implies a^2>4b$
$implies frac{a^2}{4}>b$
¿how do I use $b sim U(i,j)$?
probability probability-theory statistics probability-distributions normal-distribution
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
$endgroup$
$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08
add a comment |
$begingroup$
I am having some problems to find the solution required in the next exercise:
$x^2-ax+b=0$ has real and not equal solutions and that b is a positive constant that such that $b sim U(i,j)$ for certain i,j.
Find the $mathbb{E}[X], ,Var[X]$ of the solutions.
What I have so far:
From the Quadratic Formula, it is known that:
And since these are real solutions,
$implies a^2>4b$
$implies frac{a^2}{4}>b$
¿how do I use $b sim U(i,j)$?
probability probability-theory statistics probability-distributions normal-distribution
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
$endgroup$
I am having some problems to find the solution required in the next exercise:
$x^2-ax+b=0$ has real and not equal solutions and that b is a positive constant that such that $b sim U(i,j)$ for certain i,j.
Find the $mathbb{E}[X], ,Var[X]$ of the solutions.
What I have so far:
From the Quadratic Formula, it is known that:
And since these are real solutions,
$implies a^2>4b$
$implies frac{a^2}{4}>b$
¿how do I use $b sim U(i,j)$?
probability probability-theory statistics probability-distributions normal-distribution
probability probability-theory statistics probability-distributions normal-distribution
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
edited Dec 6 '18 at 3:27
Israel Barquín
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
asked Dec 5 '18 at 23:03
Israel BarquínIsrael Barquín
276
276
$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08
add a comment |
$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08
$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08
$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08
add a comment |
1 Answer
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$begingroup$
Well, simply by using the fact that since $b sim U(0,a^2/4)$ then $mathbb{E}(b) =4/a^2int_{0}^{a^2/2}b,db =frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$mathbb{E}(x_1) = mathbb{E}left(frac{a + sqrt{a^2-4b}}{2}right) = frac{1}{2}mathbb{E}left(a+sqrt{a^2-4b}right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $mathbb{E}(a) = a$ and so :
$$mathbb{E}(x_1) = frac{1}{2}a + frac{1}{2}mathbb{E}left(sqrt{a^2-4b}right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b sim U(0,a^2/4)$ then :
$$V(b) = frac{1}{12}left(frac{a^2}{4}-0right)^2= frac{a^4}{12cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.
edited Dec 6 '18 at 2:21
Israel Barquín
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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active
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$begingroup$
Well, simply by using the fact that since $b sim U(0,a^2/4)$ then $mathbb{E}(b) =4/a^2int_{0}^{a^2/2}b,db =frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$mathbb{E}(x_1) = mathbb{E}left(frac{a + sqrt{a^2-4b}}{2}right) = frac{1}{2}mathbb{E}left(a+sqrt{a^2-4b}right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $mathbb{E}(a) = a$ and so :
$$mathbb{E}(x_1) = frac{1}{2}a + frac{1}{2}mathbb{E}left(sqrt{a^2-4b}right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b sim U(0,a^2/4)$ then :
$$V(b) = frac{1}{12}left(frac{a^2}{4}-0right)^2= frac{a^4}{12cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.
edited Dec 6 '18 at 2:21
Israel Barquín
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
$endgroup$
add a comment |
$begingroup$
Well, simply by using the fact that since $b sim U(0,a^2/4)$ then $mathbb{E}(b) =4/a^2int_{0}^{a^2/2}b,db =frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$mathbb{E}(x_1) = mathbb{E}left(frac{a + sqrt{a^2-4b}}{2}right) = frac{1}{2}mathbb{E}left(a+sqrt{a^2-4b}right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $mathbb{E}(a) = a$ and so :
$$mathbb{E}(x_1) = frac{1}{2}a + frac{1}{2}mathbb{E}left(sqrt{a^2-4b}right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b sim U(0,a^2/4)$ then :
$$V(b) = frac{1}{12}left(frac{a^2}{4}-0right)^2= frac{a^4}{12cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.
edited Dec 6 '18 at 2:21
Israel Barquín
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
$endgroup$
add a comment |
$begingroup$
Well, simply by using the fact that since $b sim U(0,a^2/4)$ then $mathbb{E}(b) =4/a^2int_{0}^{a^2/2}b,db =frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$mathbb{E}(x_1) = mathbb{E}left(frac{a + sqrt{a^2-4b}}{2}right) = frac{1}{2}mathbb{E}left(a+sqrt{a^2-4b}right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $mathbb{E}(a) = a$ and so :
$$mathbb{E}(x_1) = frac{1}{2}a + frac{1}{2}mathbb{E}left(sqrt{a^2-4b}right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b sim U(0,a^2/4)$ then :
$$V(b) = frac{1}{12}left(frac{a^2}{4}-0right)^2= frac{a^4}{12cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.
edited Dec 6 '18 at 2:21
Israel Barquín
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
$endgroup$
Well, simply by using the fact that since $b sim U(0,a^2/4)$ then $mathbb{E}(b) =4/a^2int_{0}^{a^2/2}b,db =frac{a^2/4}{2} = a^2/8$.
Then, for $x_1$ for example, it is :
$$mathbb{E}(x_1) = mathbb{E}left(frac{a + sqrt{a^2-4b}}{2}right) = frac{1}{2}mathbb{E}left(a+sqrt{a^2-4b}right)$$
But, $a$ is a constant (not following any distributions, as far as one can say per the problem hypothesis), thus $mathbb{E}(a) = a$ and so :
$$mathbb{E}(x_1) = frac{1}{2}a + frac{1}{2}mathbb{E}left(sqrt{a^2-4b}right) $$
Can you finish the calculation now ?
For the calculation of variance, again use the fact that since $b sim U(0,a^2/4)$ then :
$$V(b) = frac{1}{12}left(frac{a^2}{4}-0right)^2= frac{a^4}{12cdot 16}$$
Be careful now, since if $a$ is a constant, then $V(a) = a^2$.
edited Dec 6 '18 at 2:21
Israel Barquín
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
edited Dec 6 '18 at 2:21
Israel Barquín
276
edited Dec 6 '18 at 2:21
Israel Barquín
276
edited Dec 6 '18 at 2:21
Israel Barquín
276
276
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
answered Dec 5 '18 at 23:16
RebellosRebellos
14.6k31247
14.6k31247
add a comment |
add a comment |
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$begingroup$
Please do not deface the question after it's answered.
$endgroup$
– Saad
Dec 6 '18 at 3:08