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Thread: Finding probability density, mean and variance of Y=2X+3

  1. #1
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    Finding probability density, mean and variance of Y=2X+3

    Suppose X has a probability density given by:

    $\displaystyle f(x)=\left\{\begin{array}{cc} \alpha e^{-\alpha x},&\mbox{for}\ x\geq 0\\0 & \mbox{ otherwise}\end{array}\right.$

    1) Find the probability density of $\displaystyle Y=2X+3$

    2) Find the mean and variance of $\displaystyle Y=2X+3$

    Could someone please explain or refer me to someone website which explains the 'scale effects'. I tried reading it in my textbook, but don't really understand it.
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  2. #2
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    by scale effects do you mean you want an explanation of why

    $\displaystyle Var(aX+b) = a^2 Var(X)$
    and
    $\displaystyle E(aX+b) = aE(x) + b$

    ?
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  3. #3
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    Quote Originally Posted by SpringFan25 View Post
    by scale effects do you mean you want an explanation of why

    $\displaystyle Var(aX+b) = a^2 Var(X)$
    and
    $\displaystyle E(aX+b) = aE(x) + b$

    ?
    Yeh kind of. I'll give you an example

    Say X is a continuous random variable with PDF $\displaystyle f(x)$ and CDF $\displaystyle F(x)$. Suppose $\displaystyle Y=aX+b$ where $\displaystyle a \ne 0$ has PDF $\displaystyle g(x)$ and CDF $\displaystyle G(x)$.

    What I don't understand is the meaning of $\displaystyle F(x)=G(ax+b)$ and $\displaystyle g(y)=\frac{1}{a}f\left(\frac{y-b}{a}\right)$ for $\displaystyle a>0$
    Last edited by acevipa; Jun 6th 2010 at 05:02 AM.
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  4. #4
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    Ok, lets examine the CDF of Y.

    Step 1:
    $\displaystyle G(Y) = P(Y<y) = P(aX + b < y)$
    $\displaystyle G(Y) = P(Y<y) = P(X < \frac{y - b}{a})$
    $\displaystyle G(Y) = P(X < \frac{y - b}{a}) = F(\frac{y-b}{a})$


    Step 2:
    Now, to get any pdf from a cdf, we just differenciate.
    $\displaystyle g(y) = \frac{dG}{dy} = \frac{dF \left( \frac{y-b}{a} \right)}{dy}$

    You can do this derivative easily using the chain rule

    $\displaystyle g(y) = \frac{dF \left( \frac{y-b}{a} \right)}{dy} = \frac{1}{a} \times F' \left( \frac{y-b}{a} \right)$
    Notation point: $\displaystyle F' \left( \frac{y-b}{a} \right)$ means the derivative of F with respect to x, evaluated at the point $\displaystyle \left( \frac{y-b}{a} \right)$

    But by definition, the derivative of F is the pdf (ie, F' = f).
    $\displaystyle g(y) = \frac{1}{a} \times f \left( \frac{y-b}{a} \right)$





    To explain why F(X) = G(aX + b) you can follow the argument in step 1 again, but try to find F() instead of G(). I've done it as a spoiler in case you get stuck

    Spoiler:

    $\displaystyle X=\frac{Y-3}{2}$

    Lets examine the PDF of X
    $\displaystyle F(X) = P(X<x) = P(\frac{Y-3}{2}) < x) $
    $\displaystyle F(X) = P(X<x) = P(Y < 2x + 3)$
    $\displaystyle F(X) = P(Y < 2x + 3) = G(2x + 3)$




    Last edited by SpringFan25; Jun 6th 2010 at 05:37 AM.
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  5. #5
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    Quote Originally Posted by SpringFan25 View Post
    Ok, lets examine the CDF of Y.

    Step 1:
    $\displaystyle G(Y) = P(Y<y) = P(aX + b < y)$
    $\displaystyle G(Y) = P(Y<y) = P(X < \frac{y - b}{a})$
    $\displaystyle G(Y) = P(X < \frac{y - b}{a}) = F(\frac{y-b}{a})$


    Step 2:
    Now, to get any pdf from a cdf, we just differenciate.
    $\displaystyle g(y) = \frac{dG}{dy} = \frac{dF \left( \frac{y-b}{a} \right)}{dy}$

    You can do this derivative easily using the chain rule

    $\displaystyle g(y) = \frac{dF \left( \frac{y-b}{a} \right)}{dy} = \frac{1}{a} \times F' \left( \frac{y-b}{a} \right)$
    Notation point: $\displaystyle F' \left( \frac{y-b}{a} \right)$ means the derivative of F with respect to x, evaluated at the point $\displaystyle \left( \frac{y-b}{a} \right)$

    But by definition, the derivative of F is the pdf (ie, F' = f).
    $\displaystyle g(y) = \frac{1}{a} \times f \left( \frac{y-b}{a} \right)$





    To explain why F(X) = G(aX + b) you can follow the argument in step 1 again, but try to find F() instead of G(). I've done it as a spoiler in case you get stuck

    Spoiler:

    $\displaystyle X=\frac{Y-3}{2}$

    Lets examine the PDF of X
    $\displaystyle F(X) = P(X<x) = P(\frac{Y-3}{2}) < x) $
    $\displaystyle F(X) = P(X<x) = P(Y < 2x + 3)$
    $\displaystyle F(X) = P(Y < 2x + 3) = G(2x + 3)$





    Thanks, I kind of understand that now. Although how do you approach the question. Do you just remember the formula??
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  6. #6
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    I never saw that formula before actually


    The principle i remembered was "Find the CDF of Y and differenciate to get the PDF"
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  7. #7
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    Quote Originally Posted by SpringFan25 View Post
    I never saw that formula before actually


    The principle i remembered was "Find the CDF of Y and differenciate to get the PDF"

    In my case, do I integrate first to find F(x) then use that to find G(Y) then F(Y) then f(y)??
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  8. #8
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    Im not clear on what you mean, there are 2 ways to get the pdf of Y.


    Short way:
    just plug the numbers into the formula we just derived


    Long way:
    Step 1:
    Find the CDF of X

    HintYou can either integrate, or notice that we are dealing with the pdf of an exponential distribution. You can lookup the CDF of an exponential distribution

    Step 2:
    Find the CDF of Y (use the same reasoning i did in the derivation above)

    Step 3:
    Differentiate the CDF of Y to get the pdf of Y.
    Last edited by SpringFan25; Jun 6th 2010 at 06:35 AM.
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  9. #9
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    Quote Originally Posted by SpringFan25 View Post
    Im not clear on what you mean, there are 2 ways to get the pdf of Y.


    Short way:
    just plug the numbers into the formula we just derived


    Long way:
    Step 1:
    Find the CDF of X

    HintYou can either integrate, or notice that we are dealing with the pdf of an exponential distribution. You can lookup the CDF of an exponential distribution

    Step 2:
    Find the CDF of Y (use the same reasoning i did in the derivation above)

    Step 3:
    Differentiate the CDF of Y to get the pdf of Y.
    Thanks for that
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