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Thread: constants Of density function

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    constants Of density function

    Hi I was given this question in an exam and unfortunately didnt know how to do it obtaining zero out of seven.
    A continuous random variable, X, had density function of the form


    $\displaystyle f(x)=\left( \begin{array}{cc}0, & \mbox{ if } x<1\\a+\frac{b}{x^4}, & \mbox{ if } x>1\end{array}\right)$
    where a and b are constants(i) calculate the values of a and b.

    I figured since the integral of the density function from $\displaystyle -\infty$ to $\displaystyle +\infty$ is 1, a and b could be derived but i don't get anywhere.
    here is my working:-
    $\displaystyle \int_1^\infty(a+\frac{b}{x^4})dx=a\int_1^\infty(1) dx+b\int_1^\infty(x^-4)dx$ which gives me $\displaystyle a(\infty-1)+b(-\frac{1}{\infty}-\frac{1}{3})=1$ Im not exactly sure but i think this gives me roughly ab=1,,,do i just sub then for instance $\displaystyle a=\frac{1}{b}$back into my original integral??? im not sure as this gives me $\displaystyle b=\infty$
    If anyone could give me some idea as to how I could get a reasonable or solid answer you'll be very helpful!!!
    Last edited by oxrigby; Jan 27th 2009 at 04:24 PM.
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    Quote Originally Posted by oxrigby View Post
    Hi I was given this question in an exam and unfortunately didnt know how to do it obtaining zero out of seven.
    A continuous random variable, X, had density function of the form


    $\displaystyle f(x)=\left( \begin{array}{cc}0, & \mbox{ if } x<1\\a+\frac{b}{x^4}, & \mbox{ if } x>1\end{array}\right)$
    where a and b are constants(i) calculate the values of a and b.

    I figured since the integral of the density function from $\displaystyle -\infty$ to $\displaystyle +\infty$ is 1, a and b could be derived but i don't get anywhere.
    here is my working:-
    $\displaystyle \int_1^\infty(a+\frac{b}{x^4})dx=a\int_1^\infty(1) dx+b\int_1^\infty(x^-4)dx$ which gives me $\displaystyle a(\infty-1)+b(-\frac{1}{\infty}-\frac{1}{3})=1$ Im not exactly sure but i think this gives me roughly ab=1,,,do i just sub then for instance $\displaystyle a=\frac{1}{b}$back into my original integral??? im not sure as this gives me $\displaystyle b=\infty$
    If anyone could give me some idea as to how I could get a reasonable or solid answer you'll be very helpful!!!
    Psssst ...... here's a tip: $\displaystyle a = 0$.
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    thanks but Ive only got as far now as $\displaystyle a(\infty-1)+b(\frac{1}{\infty}-1)=1$ which gives $\displaystyle a\infty-a-b=1$ do just assume at this point that a=0 ?
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    Quote Originally Posted by oxrigby View Post
    erm,,how? ive only got as far now as $\displaystyle a(\infty-1)+b(\frac{1}{\infty}-1)=1$ which gives $\displaystyle a\infty-b=1$ cant really see how i get a=0 from this
    It is impossible for the integral to equal 1 (or in fact, to even be finite) if $\displaystyle a \neq 0$. So $\displaystyle a = 0$.

    To get $\displaystyle b$ you need to treat the integral as an improper integral. That means solving $\displaystyle \lim_{\alpha \rightarrow +\infty} \left[ -\frac{b}{3x^3} \right]_1^{\alpha} = 1$ for $\displaystyle b$.

    (By the way, as you would realise from what I have posted, your integral calculations are wrong).
    Last edited by mr fantastic; Jan 27th 2009 at 08:14 PM. Reason: Had 0 rather than +oo
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    this is wrong?! $\displaystyle a\left[1\right]_1^\infty+b\left[\frac{-1}{3x^3} \right]_1^{\infty}=1$ giving $\displaystyle a(\infty-1)+b(\frac{1}{\infty}-1)=1$ and $\displaystyle a\infty-a-b=1$ hence the only way you can prove a=0 how else could you prove a=0 without this integral being correct?''. Presuming this is what ive done for the proof of a=0 what is wrong with $\displaystyle \left[\frac{-b}{3x^{3}}\right]_1^\infty=1$ Should you always use limits then even in the first stage for a?
    Last edited by mr fantastic; Jan 27th 2009 at 08:06 PM. Reason: Fixed latex
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    Quote Originally Posted by oxrigby View Post
    this is wrong?! $\displaystyle a\left[1\right]_1^\infty+b\left[\frac{-1}{3x^3} \right]_1^{\infty}=1$ giving $\displaystyle a(\infty-1)+b(\frac{1}{\infty}-1)=1$ and $\displaystyle a\infty-a-b=1$ hence the only way you can prove a=0 how else could you prove a=0 without this integral being correct?''. Presuming this is what ive done for the proof of a=0 what is wrong with $\displaystyle \left[\frac{-b}{3x^{3}}\right]_1^\infty=1$ Should you always use limits then even in the first stage for a?
    Infinity is not a number. You cannot substitute it into an expression. You have to use a limiting process. I assumed that you have been taught about improper integrals. Go back and review that material. There are numerous mistakes you're making with the integration - you should review that material too.
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    Quote Originally Posted by mr fantastic View Post
    Infinity is not a number. You cannot substitute it into an expression. You have to use a limiting process. I assumed that you have been taught about improper integrals. Go back and review that material. There are numerous mistakes you're making with the integration - you should review that material too.
    Note:

    $\displaystyle \lim_{\alpha \rightarrow + \infty} \left[ -\frac{b}{3x^3} \right]_1^{\alpha} = 1$

    $\displaystyle \Rightarrow \lim_{\alpha \rightarrow + \infty} \left(-\frac{b}{3 \alpha^3} + \frac{b}{3} \right) = 1$

    $\displaystyle \Rightarrow \frac{b}{3} = 1 \Rightarrow b = 3$.
    Last edited by mr fantastic; Jan 27th 2009 at 08:23 PM.
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    that makes sense! erm I know how to use limits for f(x)/g(x) i.e both polynomials. by doing the limit to 0 and replacing x by 1/y. Or l'hopital ifthey both equal zero when you put zero in f(x) and g(x) but is not this question just got a constant on the top? i dont know how to do this one.
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    Quote Originally Posted by oxrigby View Post
    that makes sense! erm I know how to use limits for f(x)/g(x) i.e both polynomials. by doing the limit to 0 and replacing x by 1/y. Or l'hopital ifthey both equal zero when you put zero in f(x) and g(x) but is not this question just got a constant on the top? i dont know how to do this one.
    Note: $\displaystyle \lim_{\alpha \rightarrow +\infty} \frac{1}{\alpha^3} = 0$.

    (There was an obvious typo which I fixed. I had 0 instead of $\displaystyle \infty$.
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