# Var(ax^2 - x )

1. Aug 28, 2014

### somecelxis

1. The problem statement, all variables and given/known data

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 .

2. Relevant equations

3. The attempt at a solution

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Last edited: Aug 28, 2014
2. Aug 28, 2014

### ehild

What is h in the pdf?
How do you get E(x) and Var(x)?

ehild

3. Aug 28, 2014

### Ray Vickson

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. Aug 28, 2014

### somecelxis

sorry. i can only post the thumbnail for this type of question. i really dont know to type the f(t) in 2 row using normal input method.

5. Aug 28, 2014

### Ray Vickson

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:
$$f(x) = \begin{cases} f_1(x), & x \leq a\\ f_2(x), & a < x \leq b \end{cases}$$
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).

6. Aug 28, 2014

### somecelxis

\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. Aug 28, 2014

### somecelxis

h = 0.2

8. Aug 28, 2014

### Ray Vickson

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:
$$E(3 X^2-X)^2 = \frac{1}{5} \int_0^5 (3 x^2-x)^2 \, dx = 5675/12$$
Thus uses the so-called "Law of the Unconscious Statistician", which says that
$$E h(X) = \int f_X(x) h(x) \, dx$$
See, eg., http://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician .