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Math Help - Differentiating a cumulative distribution function

  1. #1
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    Differentiating a cumulative distribution function

    Obviously I'm stuck. The question is to find the density function, Y = 2X.

    Fy(a) = Pr{Y<=a}
    = Pr{2X<=a}
    = Pr{X<=a/2}
    = Fx(a/2)

    Differentiate both Fy and Fx cdfs

    fy(a) = 1/2fx(a/2)

    I don't understand where the 1/2 comes from. Differentiating a cdf like F(a) results in the density function f(a), so wouldn't differentiating Fx(a/2) produce fx(a/2)?
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    Quote Originally Posted by notgoodatmath View Post
    Obviously I'm stuck. The question is to find the density function, Y = 2X.

    Fy(a) = Pr{Y<=a}
    = Pr{2X<=a}
    = Pr{X<=a/2}
    = Fx(a/2)

    Differentiate both Fy and Fx cdfs

    fy(a) = 1/2fx(a/2)

    I don't understand where the 1/2 comes from. Differentiating a cdf like F(a) results in the density function f(a), so wouldn't differentiating Fx(a/2) produce fx(a/2)?
    Use the chain rule. Let u = a/2.

    Then \frac{dF_x(a/2)}{da} = \frac{dF_x(u)}{du} \cdot \frac{du}{da} = f_x(u) \cdot \frac{1}{2} = f_x(a/2) \cdot \frac{1}{2}.
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    Thanks! I got it now.

    I'm also looking for the distribution function of Y with Y=e ^x

    F_y(x) = P{Y<=x}
    = P{e ^x <= x}
    = P{X <= log(x)}
    = \int^{log(x)}_0f(y)d(y)
    = log(x)

    I'm getting stuck at the integral stage. Could you explain how to get from the integral to the answer of log(x)? Isn't it just f(log(x))-0 = f(log(x))?
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    Quote Originally Posted by notgoodatmath View Post
    Thanks! I got it now.

    I'm also looking for the distribution function of Y with Y=e ^x

    F_y(x) = P{Y<=x} Mr F says: This notation is poor. You should use notation like F(y) = Pr(Y < y) = Pr(e^X < y) ....

    = P{e ^x <= x}
    = P{X <= log(x)}
    = \int^{log(x)}_0f(y)d(y)
    = log(x)

    I'm getting stuck at the integral stage. Could you explain how to get from the integral to the answer of log(x)? Isn't it just f(log(x))-0 = f(log(x))?
    You need to say what distribution X follows before any integration can be done to get the cdf for Y.
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    Ok. I have a density function for X: f(x) = \{^{1 if 0 \le x \le 1}_{0 otherwise}. The question asks me to find E[e ^x].

    To simplify, Y = e ^x. To determine Fy (cdf of Y), for 1 \le x \le e

    Fy = Pr(Y \le x)

    = Pr(e ^x \le log(x))

    = Pr( X \le log(x))

    = \int^{log(x)}_0f(y)dy **My query is at this stage**

    = log(x)

    Then, differentiate Fy so that fy = \frac{1}{x}(when 1 \le x \le e )

    After a bit more integration the answer comes to e - 1. I don't get how the integral marked with an asterisk is calculated though.
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    Quote Originally Posted by notgoodatmath View Post
    Ok. I have a density function for X: f(x) = \{^{1 if 0 \le x \le 1}_{0 otherwise}. The question asks me to find E[e ^x].

    To simplify, Y = e ^x. To determine Fy (cdf of Y), for 1 \le x \le e

    Fy = Pr(Y \le x)

    = Pr(e ^x \le log(x))

    = Pr( X \le log(x))

    = \int^{log(x)}_0f(y)dy **My query is at this stage**

    = log(x)

    Then, differentiate Fy so that fy = \frac{1}{x}(when 1 \le x \le e )

    After a bit more integration the answer comes to e - 1. I don't get how the integral marked with an asterisk is calculated though.
    I see you didn't take my advice regarding notation - thta's probably part of your trouble.

    F(y) = \Pr(Y \leq y) = \Pr(e^X \leq y) = \Pr(X \leq \ln y) = \int_{0}^{\ln y} 1 \, dx = [x]_{0}^{\ln y} = \ln y for 1 \leq y \leq e.

    (Note that 0 \leq X \leq 1 \Rightarrow 1 \leq Y = e^X \leq e).

    Then the pdf for Y is f(y) = \frac{dF}{dy} = \frac{1}{y} for 1 \leq y \leq e and zero elsewhere.

    Then E(e^X) = E(Y) = \int_1^e y \, \left(\frac{1}{y}\right) \, dy = e - 1.

    -----------------------------------------------------------------------------

    But why do all this when you can just use E[g(X)] = \int_{-\infty}^{+\infty} g(x) \, f(x) \, dx where f(x) is the pdf of X ......!?

    In your case E(e^X) = \int_0^1 e^x (1) \, dx = e - 1. Much less effort!
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    Quote Originally Posted by mr fantastic View Post
    I see you didn't take my advice regarding notation - thta's probably part of your trouble.

    F(y) = \Pr(Y \leq y) = \Pr(e^X \leq y) = \Pr(X \leq \ln y) = \int_{0}^{\ln y} 1 \, dx = [x]_{0}^{\ln y} = \ln y for 1 \leq y \leq e.

    (Note that 0 \leq X \leq 1 \Rightarrow 1 \leq Y = e^X \leq e).

    Then the pdf for Y is f(y) = \frac{dF}{dy} = \frac{1}{y} for 1 \leq y \leq e and zero elsewhere.

    Then E(e^X) = E(Y) = \int_1^e y \, \left(\frac{1}{y}\right) \, dy = e - 1.

    -----------------------------------------------------------------------------

    But why do all this when you can just use E[g(X)] = \int_{-\infty}^{+\infty} g(x) \, f(x) \, dx where f(x) is the pdf of X ......!?

    In your case E(e^X) = \int_0^1 e^x (1) \, dx = e - 1. Much less effort!
    Perfect. That's just what I wanted. Thanks!

    I was just trying to understand the derivation. The notation that I used was as in the probability book I'm using - perhaps I should dump it? I just copied out the notation there in case I made a mistake. But I agree, it's simpler the way you put it there. cheers!
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