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Math Help - Normal distribution?

  1. #1
    Member courteous's Avatar
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    Question Normal distribution?

    X\sim N(\mu,\sigma).
    Y\equiv aX+b.
    What is Y's distribution?
    I guess Y is still normally distributed, but what are \mu_Y and \sigma_Y?

    E.g. for \sigma: it's "intuitive" to think \sigma_Y = |a|\sigma_X, but how do you get the answer formally, with procedure? Is it via probability density function p_X(x)?
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    Quote Originally Posted by courteous View Post
    I guess Y is still normally distributed, but what are \mu_Y and \sigma_Y?

    E.g. for \sigma: it's "intuitive" to think \sigma_Y = |a|\sigma_X, but how do you get the answer formally, with procedure? Is it via probability density function p_X(x)?
    Usually cumulative distributive function is used to compute distributions of transformation of random variables. You can also use the Jacobian approach.

    I will start you off using the c.d.f F_{Y}(y), where y is a real number:

    F_{Y}(y) = \text{Pr}(Y \leq y) = \text{Pr}(aX+b \leq y)

    What will you do next?
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  3. #3
    Member courteous's Avatar
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    Unhappy

    I can only manage 2 steps: =Pr(X\leq \frac{y-b}{a})=\frac{1}{2}+\Phi(\frac{y-b}{a})
    Last edited by courteous; August 27th 2010 at 12:00 PM. Reason: Pr
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  4. #4
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    Quote Originally Posted by courteous View Post
    I can only manage 2 steps: =Pr(X\leq \frac{y-b}{a})=\frac{1}{2}+\Phi(\frac{y-b}{a})
    Good! So how is a c.d.f related to a p.d.f?
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  5. #5
    Member courteous's Avatar
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    Well, CDF is the sum of all the PDFs: cdf(x)=\int_{-\infty}^x pdf(x)dx.
    I'm also looking at Wikipedia on change of variables, but I don't know how to make use of it.

    I'm also guessing I shouldn't be doing \int_{-\infty}^x \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}?
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  6. #6
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    Quote Originally Posted by courteous View Post
    Well, CDF is the sum of all the PDFs: cdf(x)=\int_{-\infty}^x pdf(x)dx.
    I'm also looking at Wikipedia on change of variables, but I don't know how to make use of it.

    I'm also guessing I shouldn't be doing \int_{-\infty}^x \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}?
    You should know that the derivative of the cdf gives the pdf.
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  7. #7
    Member courteous's Avatar
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    p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{  \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...

    Now what?
    How to relate \frac{y-b}{a} with t?
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  8. #8
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    Quote Originally Posted by courteous View Post
    p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{  \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...

    Now what?
    How to relate \frac{y-b}{a} with t?
    Actually now this question belongs to the Calculus forum :P

    Repeating mr fantastic's words: You should know that (F_X(x))' = f(x) where F_X(x) is the c.d.f and f(x) is the p.d.f of a random variable X.

    So to solve the problem, I am expecting you to know the p.d.f of a normal distribution and chain rule of differentiation. Do you know these?
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  9. #9
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    Quote Originally Posted by courteous View Post
    p_Y(y)=F_Y(y)'=\big(\frac{1}{2}+\Phi(\frac{y-b}{a})\big)'=0+\big(\frac{1}{2\pi}\larger{\int_0^{  \frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'=\frac{1}{2\pi}...

    Now what?
    How to relate \frac{y-b}{a} with t?
    The derivative is with respect to y. Use the chain rule and the Fundamental Theorem of Calculus to get it.
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  10. #10
    Member courteous's Avatar
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    Isn't what's inside the integral (that I wrote above) "the p.d.f of a normal distribution"?
    Notation: I've mistakenly used p_Y(y) for f(y).
    Chain rule is \frac{da}{db}=\frac{da}{dc}\frac{dc}{db}.

    I haven't come upon such a problem before, so I'm not "used to it", as von Neumann would say.

    Was I even on the right track in my previous post?
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  11. #11
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    Quote Originally Posted by courteous View Post
    Isn't what's inside the integral (that I wrote above) "the p.d.f of a normal distribution"?
    Notation: I've mistakenly used p_Y(y) for f(y).
    Chain rule is \frac{da}{db}=\frac{da}{dc}\frac{dc}{db}.

    I haven't come upon such a problem before, so I'm not "used to it", as von Neumann would say.

    Was I even on the right track in my previous post?
    You are on the right track. However you are not using the fact that derivative of a c.d.f is the p.d.f. Your integral along with the \frac12 term is the c.d.f of a normal distribution, i.e.

    F_Y(y) = F_X\left(\frac{y-b}{a}\right) \implies p_Y(y) = (F_Y(y))' =\left(F_X\left(\frac{y-b}{a}\right)\right)' = \text{???}

    Alternatively,

    \big(\frac{1}{2\pi}\larger{\int_0^{\frac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)'  = \dfrac{d}{dy}\big(\frac{1}{2\pi}\larger{\int_0^{\f  rac{y-b}{a}}}e^{-\frac{t^2}{2}}dt\big)

    I was telling you how to proceed. You can use the fundamental theorem of calculus to differentiate integrals with respect to a variable that appears in the limit of the integral. In the above expression 'y' appears as a limit of integration. So you have to use the fundamental theorem of calculus.
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  12. #12
    Member courteous's Avatar
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    Quote Originally Posted by Isomorphism View Post
    F_Y(y) = F_X\left(\frac{y-b}{a}\right) \implies p_Y(y) = (F_Y(y))' =\left(F_X\left(\frac{y-b}{a}\right)\right)' = \text{???}
    \displaystyle{  =\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)=\frac{1}{\sigma\sqrt{2\  pi}}\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big) \overbrace{=}^{A} \frac{1}{\sigma\sqrt{2\pi}} \frac{d}{dy}\big(h(g(y))\big)}=  }


    \displaystyle{  = \frac{1}{\sigma\sqrt{2\pi}} h'\big(g(y)\big)g'(y) \overbrace{=}^{B} \frac{1}{\sigma\sqrt{2\pi}}  e^{-\frac{(\frac{y-b}{a}-\mu)^2}{2\sigma^2}} \frac{1}{a} = \frac{1}{(a\sigma)\sqrt{2\pi}}  e^{-\frac{(y-(a\mu+b))^2}{2(a\sigma)^2}}}


    \implies Y\sim N(a\mu +b, |a|\sigma) Is everything correct?




    A: Let h(y)=\int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx and g(y)=\frac{y-b}{a}.

    B: h'(y)=\frac{d}{dy}\big(h(y)\big)=\frac{d}{dy}\big(  \int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)\overbrace{=}^{FTC} e^{-\frac{(y-\mu)^2}{2\sigma^2}}. FTC = Fundamental theorem of calculus.
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  13. #13
    Lord of certain Rings
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    Quote Originally Posted by courteous View Post
    \displaystyle{  =\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)=\frac{1}{\sigma\sqrt{2\  pi}}\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big) \overbrace{=}^{A} \frac{1}{\sigma\sqrt{2\pi}} \frac{d}{dy}\big(h(g(y))\big)}=  }


    \displaystyle{  = \frac{1}{\sigma\sqrt{2\pi}} h'\big(g(y)\big)g'(y) \overbrace{=}^{B} \frac{1}{\sigma\sqrt{2\pi}}  e^{-\frac{(\frac{y-b}{a}-\mu)^2}{2\sigma^2}} \frac{1}{a} = \frac{1}{(a\sigma)\sqrt{2\pi}}  e^{-\frac{(y-(a\mu+b))^2}{2(a\sigma)^2}}}


    \implies Y\sim N(a\mu +b, |a|\sigma) Is everything correct?




    A: Let h(y)=\int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx and g(y)=\frac{y-b}{a}.

    B: h'(y)=\frac{d}{dy}\big(h(y)\big)=\frac{d}{dy}\big(  \int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)\overbrace{=}^{FTC} e^{-\frac{(y-\mu)^2}{2\sigma^2}}. FTC = Fundamental theorem of calculus.

    Perrrrfect!!
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  14. #14
    Member courteous's Avatar
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    Thank you both, Isomorphism and Mr. Fantastic, for your guidance (and patience) ... I am especially grateful that you didn't show the whole solution at once.
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  15. #15
    Member courteous's Avatar
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    The Jacobian approach

    X\sim N(\mu,\sigma).
    Y\equiv aX+b.
    What is Y's distribution?
    Quote Originally Posted by Isomorphism View Post
    Usually cumulative distributive function is used to compute distributions of transformation of random variables. You can also use the Jacobian approach.
    Isomorphism, would you show me "the Jacobian approach", please? (Other members are welcomed as well!)
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