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  1. #1
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    distribution function

    Hi can you help me to solve the following problem?. thank you

    Let X1 and X2 denote a random sample of size 2 from a distribution with p.d.f. (densitity function) f(x)=1 , 0<x<1, zero elsewhere. Find the distribution function and p.d.f. of Y=X1/X2
    Last edited by user; April 3rd 2010 at 08:46 AM.
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  2. #2
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    Quote Originally Posted by user View Post
    Hi can you help me to solve the following problem?. thank you

    Let X1 and X2 denote a random sample of size 2 from a distribution with p.d.f. (densitity function) f(x)=1 , 0<x<1, zero elsewhere. Find the distribution function and p.d.f. of Y=X1/X2
    Read these:

    Ratio Distribution -- from Wolfram MathWorld

    Ratio distribution - Wikipedia, the free encyclopedia

    Quotient of two random variables
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  3. #3
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    Again

    Hi can you help me to solve the following problem?. But without stochastics theory. I must to use theory of probability and transformations. Thank you

    Let X1 and X2 denote a random sample of size 2 from a distribution with p.d.f. (densitity function) f(x)=1 , 0<x<1, zero elsewhere. Find the distribution function and p.d.f. of Y=X1/X2
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  4. #4
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    Quote Originally Posted by user View Post
    Hi can you help me to solve the following problem?. But without stochastics theory. I must to use theory of probability and transformations. Thank you

    Let X1 and X2 denote a random sample of size 2 from a distribution with p.d.f. (densitity function) f(x)=1 , 0<x<1, zero elsewhere. Find the distribution function and p.d.f. of Y=X1/X2
    If you read the links I gave you will know that the distribution of Y/X where X and Y are continuous independent random variables with pdf's f(x) and g(y) respectively is given by

    h(u) = \int_{-\infty}^{+\infty} |y| \, f(uy) \, g(y) \, dy. This is easily proved using the 'Change of variable (transformation)' theorem.

    For your problem, note that f(uy) = 1 if 0 \leq 0 uy \leq 1 \Rightarrow 0 \leq y \leq \frac{1}{u} and zero otherwise.

    Then:

    Case 1: h(u) = 0 for u < 0.

    Case 2: h(u) = \int_0^1 y \, dy for 0 \leq u \leq 1.

    Case 3: h(u) = \int_0^{1/y} y \, dy for u > 1.
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