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Math Help - Checking a continous random variable distrubtion

  1. #1
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    Checking a continous random variable distrubtion

    apparently, according to a paper mark scheme

     <br />
f(y) = \left\{ \begin{array}{l l}<br />
 \frac{1}{9} (2y^{3} + y), & 1 \leq y \leq 2 \\<br />
 & \\<br />
  0 & \text{otherwise} \end{array} \right.<br />


    Anyway im interested in \frac{1}{9}(2y^{3}+y)

    is a valid probability density function
    But if you put the upper limit 2 in you get 2 out, when probability ranges between 0 and 1.

    Thanks
    Last edited by mr fantastic; May 31st 2009 at 05:21 AM. Reason: Fixed the latex for the array
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  2. #2
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    Quote Originally Posted by thomas49th View Post
    apparently, according to a paper mark scheme

     <br />
f(y) = \left\{ \begin{array}{l l}<br />
\frac{1}{9} (2y^{3} + y), & 1 \leq y \leq 2 \\<br />
& \\<br />
0 & \text{otherwise} \end{array} \right.<br />


    Anyway im interested in \frac{1}{9}(2y^{3}+y)

    is a valid probability density function
    But if you put the upper limit 2 in you get 2 out, when probability ranges between 0 and 1.

    Thanks
    \frac{1}{9} \int_1^2 (2y^{3} + y) \, dy = 1 so I don't see what the problem is.
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  3. #3
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    I thought a PDF tells you what the probability is for a certain value you put in.
    My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

    <br /> <br />
\frac{1}{9}(2y^{3}+y)<br />

    where y is the value you want to find the probability of getting it

    BUT if you put a 2 in you get a probability greater than 1, which cant be

    Right?

    Thanks
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  4. #4
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    Hello,
    Quote Originally Posted by thomas49th View Post
    I thought a PDF tells you what the probability is for a certain value you put in.
    My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

    <br /> <br />
\frac{1}{9}(2y^{3}+y)<br />

    where y is the value you want to find the probability of getting it

    BUT if you put a 2 in you get a probability greater than 1, which cant be

    Right?

    Thanks
    I think you're misunderstanding the role of the pdf

    Two things :
    - the probability that a continuous random variable takes a specific value is 0.
    - if X has the pdf f, then :
    \mathbb{P}(a\leq X\leq b)=\int_a^b f(x)~dx, and not something like f(b)-f(a)

    A third thing... :
    A function is a pdf if : it is a positive function, and its integral over \mathbb{R} is 1.
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  5. #5
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    Quote Originally Posted by thomas49th View Post
    I thought a PDF tells you what the probability is for a certain value you put in.
    My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

    <br /> <br />
\frac{1}{9}(2y^{3}+y)<br />

    where y is the value

    BUT if you put a 2 in you get a probability greater than 1, which cant be

    Right?

    Thanks
    You have many wrong ideas.

    1. If Y is a continuous random variable, then Pr(Y = a) = 0. It only makes sense to talk about the probabilty that Y lies between two values. You do not calculate probabilities by substituting values of y into the pdf.

    2. For y \geq 1, \Pr(Y < y) = \frac{1}{9} \int_1^y (2t^{3} + t) \, dt.

    You are strongly advised to extensively review all you have learned on continuous random variables and probability density functions.
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