Help Solving Var(ax^2 - x): Negative Answer?

  • Thread starter somecelxis
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In summary: E(X^3) = integration of hx^ 3 from 0 to 3 , where h = 0.25my E(X^2) = integration of hx^ 2 from 0 to 2 , where h = 0.5 .my E(X) = 0 In summary, you have an equation for the total variation of a function, but you are missing the function itself.
  • #1
somecelxis
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Homework Statement



Var(ax^2 - x )
can anyone please help me with part iii) ? i know that VAR (x) = E(X^2)- ( E(X) )^2 ... But I get my E(B^2) = 6875/4 , whereas my E(B) = 5125/12 , where my B= 3X^2 - X, i take E(B^2)- ( E(B) )^2, MY ANS IS NEGATIVE . WHICH IS WRONG! i have attached my working in the photo .

Homework Equations





The Attempt at a Solution

 

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  • #2
What is h in the pdf?
How do you get E(x) and Var(x)? ehild
 
  • #3
somecelxis said:

Homework Statement



Var(ax^2 - x )
can anyone please help me with part iii) ? i know that VAR (x) = E(X^2)- ( E(X) )^2 ... But I get my E(B^2) = 6875/4 , whereas my E(B) = 5125/12 , where my B= 3X^2 - X, i take E(B^2)- ( E(B) )^2, MY ANS IS NEGATIVE . WHICH IS WRONG! i have attached my working in the photo .

Homework Equations





The Attempt at a Solution



You are doing it again---posting a thumbnail instead of typing out your question. I cannot read that thumbnail now. When will you ever learn? Please read the thread on "advice for posters" at the very beginning of this Forum.
 
  • #4
Ray Vickson said:
You are doing it again---posting a thumbnail instead of typing out your question. I cannot read that thumbnail now. When will you ever learn? Please read the thread on "advice for posters" at the very beginning of this Forum.

sorry. i can only post the thumbnail for this type of question. i really don't know to type the f(t) in 2 row using normal input method.
 
  • #5
somecelxis said:
sorry. i can only post the thumbnail for this type of question. i really don't know to type the f(t) in 2 row using normal input method.

Now I am on a medium allowing me to read the thumbnail. However, I cannot write a reply and refer back to the thumbnail sumultaneously; I either have to be reading your posted picture (in non-reply mode) or else replying; I cannot do both. So, I cannot give your precise function here, but if I recall correctly you have a density function in piecewise form:
[tex] f(x) = \begin{cases} f_1(x), & x \leq a\\
f_2(x), & a < x \leq b
\end{cases}[/tex]
If you don't like to use LaTeX (which, BTW, is easy and works wonderfully) you can do it in plain text like this: f(x) = f1(x) if x <= a, and fx) = f2(x) if a < x <= b. It probably takes less time to actually type it out than to photograph it, convert the output, upload it and then attach it to a message.

To see how the above is typed in LaTeX, just right-click on the expression and choose "show math as as TeX commands". The whole expression is started with "[t e x]" (no spaces) and ended by "[/t e x]" (no spaces).
 
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  • #6
Ray Vickson said:
Now I am on a medium allowing me to read the thumbnail. However, I cannot write a reply and refer back to the thumbnail sumultaneously; I either have to be reading your posted picture (in non-reply mode) or else replying; I cannot do both. So, I cannot give your precise function here, but if I recall correctly you have a density function in piecewise form:
[tex] f(x) = \begin{cases} f_1(x), & x \leq a\\
f_2(x), & a < x \leq b
\end{cases}[/tex]
If you don't like to use LaTeX (which, BTW, is easy and works wonderfully) you can do it in plain text like this: f(x) = f1(x) if x <= a, and fx) = f2(x) if a < x <= b. It probably takes less time to actually type it out than to photograph it, convert the output, upload it and then attach it to a message.

To see how the above is typed in LaTeX, just right-click on the expression and choose "show math as as TeX commands". The whole expression is started with "[t e x]" (no spaces) and ended by "[/t e x]" (no spaces).


$\displaystyle \begin{align*} \mathrm{Var}\,\left( 3X^2 - X \right) &= \mathrm{E}\,\left[ \left( 3X^2 - X \right) ^2 \right] - \left[ \mathrm{E}\,\left( 3X^2 - X \right) \right] ^2 \\ &= \mathrm{E}\,\left( 9X^4 - 6X^3 + X^2 \right) - \left[ 3\, \mathrm{E}\,\left( X^2 \right) - \mathrm{E}\,\left( X \right) \right] ^2 \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - \left\{ 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 - 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\, \left( X \right) + \left[ \mathrm{E}\,\left( X \right) \right] ^2 \right\} \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 + 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\,\left( X \right) - \left[ \mathrm{E}\,\left( X \right) \right] ^2 \end{align*}$

my E(X^4) = integration of hx^ 4 from 0 to 5 , where h = 0.2 , so my E(X^4) = 125
by using the same method , i find E(X^3 ) , E(X^3)= integration of hx^ 3 from 0 to 5 , so my E(X^3)=125/4

by substituiting E(X^4) = 125 , E(X^3)=125/4 , E(X^2)=25/3 , E(X) = 5/2 , my ans if VAR(3X^2 - X) = 5275/12 , but the ans form book is 6025/12 . What's wrong with my working?
 
  • #7
ehild said:
What is h in the pdf?
How do you get E(x) and Var(x)?


ehild

h = 0.2
 
  • #8
somecelxis said:
$\displaystyle \begin{align*} \mathrm{Var}\,\left( 3X^2 - X \right) &= \mathrm{E}\,\left[ \left( 3X^2 - X \right) ^2 \right] - \left[ \mathrm{E}\,\left( 3X^2 - X \right) \right] ^2 \\ &= \mathrm{E}\,\left( 9X^4 - 6X^3 + X^2 \right) - \left[ 3\, \mathrm{E}\,\left( X^2 \right) - \mathrm{E}\,\left( X \right) \right] ^2 \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - \left\{ 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 - 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\, \left( X \right) + \left[ \mathrm{E}\,\left( X \right) \right] ^2 \right\} \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 + 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\,\left( X \right) - \left[ \mathrm{E}\,\left( X \right) \right] ^2 \end{align*}$

my E(X^4) = integration of hx^ 4 from 0 to 5 , where h = 0.2 , so my E(X^4) = 125
by using the same method , i find E(X^3 ) , E(X^3)= integration of hx^ 3 from 0 to 5 , so my E(X^3)=125/4

by substituiting E(X^4) = 125 , E(X^3)=125/4 , E(X^2)=25/3 , E(X) = 5/2 , my ans if VAR(3X^2 - X) = 5275/12 , but the ans form book is 6025/12 . What's wrong with my working?

I get your answer of 5275/12. Here is how I (or, rather, Maple) did it:
y:=3*x^2-x;
2
y := 3 x - x
Ey:=1/5*int(y,x=0..5);
Ey := 45/2
Ey2:=1/5*int(y^2,x=0..5);
Ey2 := 5675/6
Vy:=Ey2-(Ey)^2;
Vy := 5275/12

Note: rather than splitting things up, you can do it directly:
[tex] E(3 X^2-X)^2 = \frac{1}{5} \int_0^5 (3 x^2-x)^2 \, dx = 5675/12 [/tex]
Thus uses the so-called "Law of the Unconscious Statistician", which says that
[tex] E h(X) = \int f_X(x) h(x) \, dx [/tex]
See, eg., http://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician .
 

1. What is the equation "Var(ax^2 - x)" referring to?

The equation refers to the variance of a quadratic function with a coefficient a and a variable x.

2. How do you solve for the variance of a quadratic function?

To solve for the variance, you would first expand the equation and distribute the coefficient a. Then, use the formula for variance, which is the sum of the squares of the differences between each data point and the mean, divided by the total number of data points.

3. Why is the answer for "Var(ax^2 - x)" negative?

The answer for variance can be negative if the data points are spread out in such a way that the majority of them are farther away from the mean, resulting in a negative sum of squares.

4. How can I interpret a negative variance?

A negative variance means that the data points are more spread out and have a wider range of values compared to the mean. This could indicate a high level of variability or uncertainty in the data.

5. Are there any limitations to using variance as a measure of variability?

Yes, there are limitations to using variance as a measure of variability. It can be heavily influenced by outliers and does not provide information about the direction of the data points. Additionally, it assumes a normal distribution of the data, which may not always be the case.

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