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Thread: Symmetric probability distribution and some properties

  1. #1
    Senior Member dokrbb's Avatar
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    Symmetric probability distribution and some properties

    So, I have the following, for a real valued random variable $K$ with a pdf $f$ there is a real valued mean $\mu$ s.t.

    $f(\mu +k) = f(\mu - k)$, and I need to show that for some random variables $(\mu - K)$ and $(K - \mu)$, their c.d.f. are the same.

    From what I understand, I have a pdf for a symmetric distribution around the mean.

    Now, let $Y = (\mu - K)$, then $(K - \mu) = -(\mu - K) = -Y$, and Basically, I need to show that $F(Y) = F(-Y)$.

    But, by definition of cumulative distribution function, $F(Y) = P(Y \leq y) = \int_{-\infty}^{y}f(y)dy$, while $F(-Y) = P(Y \leq -y) = \int_{-\infty}^{-y}f(-y)dy$ and from how I see it $F(-Y) < F(Y)$.

    In fact I can define $F(Y)$ in this way, $F(Y) = P(Y \leq y) = P(-\infty \leq Y \leq -y) + P(-y < Y \leq y)$

    Could you point out what I get wrong?
    Last edited by dokrbb; Mar 11th 2017 at 03:41 PM.
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  2. #2
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    Re: Symmetric probability distribution and some properties

    let's step back a sec

    $X_1 = \mu - K$

    $X_2 = K - \mu$

    $F_{X_1}(x) = P[X_1 < x] = P[\mu - K < x] = 1 - F_K(\mu - x)$

    $F_{X_2}(x) = P[X_2 < x] = F_K(\mu + x)$

    $F_{X_1} = \displaystyle{\int_{\mu-x}^\infty}~f_K(u)~du$

    $F_{X_2} = \displaystyle{\int_{-\infty}^{\mu+x}}~f_K(u)~du$

    in the first integral let

    $v=\mu-u,~dv=-du$

    $F_{X_1}(x) = \displaystyle{\int_{-\infty}^x}~f_K(\mu - v)~dv$

    In the second integral let

    $v =u-\mu,~dv = du$

    $F_{X_2}(x) = \displaystyle{\int_{-\infty}^x}~f_K(\mu+v)~dv$

    and as $f_K(\mu-v) = f_K(\mu+v)$

    $F_{X_1}(x) = F_{X_2}(x)$
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