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Math Help - Conditional distribution of continuous random variables

  1. #1
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    Conditional distribution of continuous random variables

    Hello, I have a question about conditional distributions in the continuous case.

    We are given that the joint distribution function of two continuous random variables X and Y is

    f(x,y) = \frac{3}{2}(x^2+y^2) for \;\;0 \leq x \leq 1 , 0 \leq y \leq 1

    The marginal distribution of Y can be found by integrating with respect to x, the marginal distribution of Y is

    f_Y(y) = \frac{3}{2}y^2+\frac{1}{2} for 0 \leq y \leq 1

    The book asks to find the conditional density of X given that Y =.3

    This is the answer that is listed in the book

    f_{X|Y}(x|Y =.3) = \frac{f(x,.3)}{f_Y(.3)} = \frac{\frac{3}{2}(x^2+.3^2)}{\frac{3}{2}*.3^2+.5}=  \frac{\frac{3}{2}(x^2+.09)}{.635}

    Why is the book just plugging in the value .3 for y? Shouldn't we be integrating since this is continous?
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  2. #2
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    Re: Conditional distribution of continuous random variables

    Hey Jame.

    You can think of the conditional distribution as the slice (or cross sectional) distribution where the Y value is fixed. Since you knowe the value of y, you treat it like a constant and substitute it in to get the distribution for the random variable X.

    If you think of it as a constant and X being the random variable, then you get a PDF in terms of X only with all the properties (integrates to 1, greater than zero, and so on).
    Thanks from Jame
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  3. #3
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    Re: Conditional distribution of continuous random variables

    That makes sense.

    I guess I was thrown off by the single value and the fact that the rvs were continous.

    If I was given that Y was between 0 and .3, then would I integrate?

    Thank you for assistance.
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  4. #4
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    Re: Conditional distribution of continuous random variables

    Yes you would integrate in that situation over the appropriate region.
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