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Probability density function of uniform distribution

  1. Jul 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Random variable X is uniformly distributed on interval [0,1]:
    [tex]f(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq 1\\ 0 & \text{ else} \end{cases}[/tex]

    a) Find probability density function ρ(y) of random variable [tex]Y=\sqrt{X} +1[/tex]

    I tried like this. Is it good, if no why not?

    [tex]F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}[/tex]
     
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  3. Jul 15, 2015 #2

    HallsofIvy

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    How did [itex]\sqrt{X}+ 1[/itex] become [itex]X^2+ 1[/itex]? If you intended the inverse function then it should be [itex]X= (Y- 1)^2[/itex]
     
  4. Jul 15, 2015 #3
    Hmm, i see.
    Ok then, i get this
    [tex]F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2} \\ f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |=f_{X}(y+1)^{2})*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}[/tex][/QUOTE]

    But what about y's probability density function ?
    Is it ok?
     
  5. Jul 15, 2015 #4

    haruspex

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    That should really be
    ##F_{Y}(y)=P(Y\leq y)=P(X^{\frac 12}+1\leq y)##
    Check that step.
    ##F_{X}'((y+1)^{2})## just means the derivative wrt x of FX(x) evaluated at x = (y+1)2
    ##f_{X}(x)=F_{X}'((y+1)^{2})=f_{X}((y+1)^{2})##
     
  6. Jul 18, 2015 #5

    Ray Vickson

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    But what about y's probability density function ?
    Is it ok?[/QUOTE]

    How do you get from ##x^{1/2}+1 \leq y## to ##x \leq (y+1)^2##? (I don't.)

    Anyway, since ##X## ranges from 0 to 1, so does ##\sqrt{X}## and that means that ##Y = 1 + \sqrt{X}## ranges from 1 to 2. So, you ought to have ##F_Y(1) = 0## and ##F_Y(2) = 1##. Do you have that?
     
    Last edited: Jul 18, 2015
  7. Jul 20, 2015 #6

    andrewkirk

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    Up to here it's OK, except for the two errors pointed out by haruspex. Fix those and you will have a correct formula for ##F_Y(y)## in terms of ##F_X## applied to some function of ##y##.

    Next you can use the fact that ##F_X(x)=\int_0^x f_X(u)du=\int_0^x du=x## to write a formula for ##F_{Y}(y)## in terms of ##y##.

    Forget about everything you wrote below your first line, as that heads off on a goose chase involving ##f_X## and there is no need to refer to ##f_X## any further.

    Differentiating your formula for ##F_Y(y)## will give you a formula for ##f_Y(y)##.

    To complete the answer you need to put range constraints into your expression for ##f_Y##, along the lines suggested by Ray.
     
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