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    Help with a hw question




    I am taking my first probability course and need some help. I am working on some homework. The question I am on now asks...

    Consider fX(x)=Ce^-(x^4)

    Calculate the normalization factor C which makes fX a pdf.

    I think that I should integrate from -inf to 0 and 0 to + inf. to calculate C

    Calculate the constant a>0 for which Y=aX has a standard deviation = 1.

    Not sure about this one

    What is the pdf of the random variable Y?

    not sure about this one

    What is the fourth moment of Y?

    hmmm

    the hint is to use the gamma function.

    anybody able to help guide me through this?

    thanks!

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    Re: Help with a hw question

    1) For fX to be a pdf, the integral from -inf to +inf must =1. So you integrate fX = Ce^-(x^4) from -inf to +inf and set it equal to 1. Then solve for C.

    2) Given Y=aX, then Var(Y) = a^2 * Var(X). But then you don't know Var(X)? A brute force way to find Var(X) = E(X^2) - E^2(X), and find the expectations by integrating x^2 * fX

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    Re: Help with a hw question

    Well when i integrate I get -[1/(4x^3)]*e^-(x^4). I think that I need to chop this up into two integrals. (- inf) to 0 and 0 to (inf). Therefore...

    C= -2x^3 for -inf to 0

    C= 2x^3 for 0 to inf.

    does this sound correct? Any guidance on how to incorporate the gamma function?

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    Re: Help with a hw question

    If the support of X is \mathbb{R},

    then since the pdf f_X(x) = Ce^{-x^4} is an even function,

    i.e. f_X(x) = f_X(-x) ~ \forall x \in \mathbb{R}

    You may rewrite your integral

    \int_{-\infty}^{+\infty} f_X(x) dx = 2\int_0^{+\infty} f_X(x)dx

    And use the substitution u = x^4 in the integral,
    (when x = 0, u = 0; x \to +\infty, u \to +\infty; 
dx = \frac {1} {4} u^{-\frac {3} {4}} du )
    you can convert your integral in terms of gamma integral.

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    Re: Help with a hw question

    Update

    I am taking my first probability course and need some help. I am working on some homework. The question I am on now asks...

    Consider fX(x)=Ce^-(x^4)

    Calculate the normalization factor C which makes fX a pdf.

    thanks for the help on this folks, I found C= 2/[gamma(1/4)]

    Calculate the constant a>0 for which Y=aX has a standard deviation = 1.

    I found EX to be = gamma(3/2)/[2*gamma(1/4)]

    EX^2= gamma(5/4)/[2*gamma(1/4)]

    Var X= 1/8 - 1/64 = 7/64 (used numerical values for gamma fxns from wikipedia)

    and used VarY=a^2 * VarX to solve for a which i found a= [8*sqrt(7)]/7

    What is the pdf of the random variable Y?

    therefore the pdf of Y must be 2/[gamma(1/4)] * e^(-(8^4/49)*y^4)

    What is the fourth moment of Y?

    I am getting some crazy stuff for this, using brute force to find EY,EY^2,EY^3, etc. This question seems to invite the use of the mgf in some way, but i am missing the trick to do that. Hmm maybe i should just generate the mgf with e^tx

    the hint is to use the gamma function.

    anybody able to help guide me through this?

    thanks!

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    Re: Help with a hw question


    I have not check your solution in red.

    However, I think evaluating the moments E[Y^n] here
    is not tedious; actually it use the same trick, same substitution and
    use the gamma integral again.

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