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Math Help - Weibull & Exponential distribution

  1. #1
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    Weibull & Exponential distribution

    X is a random variable with weibull distribution given with

    <br />
f(x;\alpha ,\beta) = \begin{cases}<br />
\frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}} & x>0\\<br />
0 & x=0\end{cases}<br />

    Let \alpha=2 and  Y=2X^2/\beta^2.

    Show that Y has exponential distribution and find value of parameter \lambda.

    I believe that G(Y)=P(Y \leq y) should be expressed in terms of X , but I don't know where to start.
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  2. #2
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    you have the right approach

    Lets find P(Y<y)
    P(Y \leq y) = P(2x^2/B^2 \leq y) = P(x \leq \sqrt {0.5 B^2 y}) = F(\sqrt {0.5 B^2 y})

    Find (or look up) F(x) then subsititute these values in to get the CDF of Y. Hopefully it will look like an exponential distribution.
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  3. #3
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    Quote Originally Posted by losm1 View Post
    X is a random variable with weibull distribution given with

    <br />
f(x;\alpha ,\beta) = \begin{cases}<br />
\frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}} & x>0\\<br />
0 & x=0\end{cases}<br />

    Let \alpha=2 and  Y=2X^2/\beta^2.

    Show that Y has exponential distribution and find value of parameter \lambda.

    I believe that G(Y)=P(Y \leq y) should be expressed in terms of X , but I don't know where to start.
    Since you know in advance that the answer is going to be a well-known distribution, a simple approach would be to calculate the moment generating function of Y ....


    But if you're determined to use a distribution function approach, then you should start by noting that the cdf of Y is:

    F(y) = \Pr(Y \leq y) = \Pr \left( \frac{2 X^2}{\beta^2} \leq y \right) = \Pr \left( - \frac{\beta}{\sqrt{2}} \sqrt{y} \leq X \leq \frac{\beta}{\sqrt{2}} \sqrt{y} \right)

     =  \Pr \left( 0 < X \leq \frac{\beta}{\sqrt{2}} \sqrt{y} \right) since the support of X is x \geq 0.

    Now set up the required integral (no need to integrate by the way ....) and then differentiate using the chain rule and the Fundamental Theorem of Calculus.
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  4. #4
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    Quote Originally Posted by SpringFan25 View Post
    you have the right approach

    Lets find P(Y<y)
    P(Y \leq y) = P(2x^2/B^2 \leq y) = P(x \leq \sqrt {0.5 B^2 y}) = F(\sqrt {0.5 B^2 y})

    Find (or look up) F(x) then subsititute these values in to get the CDF of Y. Hopefully it will look like an exponential distribution.
    Thanks, I just plugged in your result in weibull CDF and got exponential CDF with \lambda=0.5

    I hope that it's correct.
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