# Thread: Symmetric probability distribution and some properties

1. ## 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?

2. ## 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 $f_K(\mu-v) = f_K(\mu+v)$

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