This is what i know about symmetric distributions: The distribution of rv (random variable) $X$ is symmetric about $a$ iff $$ P ( X \le a - x ) = P ( X \ge a + x ) \qquad \forall x \in \mathbb{R} $$ I just want to confirm ,.,is this " iff " correct ? i mean it is clear to me that if $x$ is symmetric about $a$ it will satisfy above equation, is the converse also true ?


A (real) random variable $X$ is symmetric about zero iff $X$ and $-X$ have the same distribution, often written as $\newcommand{\eqD}{\stackrel{\small{D}}{=}} X \eqD -X$. Now if $X$ is symmetric about $a$, $X-a$ will be symmetric about 0, so we have $X-a \eqD a-X$. If $X$ has a density, in terms of density functions this becomes $f(x) = f(-x)$, or, for symmetry about $a$, $f(a+x)=f(a-x)$ so the density function is symmetric about $a$.

Now, we can use this definition to calculate the cumulative distribution function $F(x)=P(X \le x)$ in two different ways: \begin{align} P(X-a \le x) &= P(-X+a \le x) \\ P(X \le x+a) &= P(-X \le x-a) \\ P(X \le x+a) &= 1-P(X \le a-x) \\ F(x+a) &= 1-F(a-x) \end{align} and this is almost what you have given, but you need to switch your second inequality sign! And, yes, this is an equivalence.

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    $\begingroup$ Your equating of $P(-X\leq x-a)$ with $1-P(X\leq a-x)$ is valid only for continuous distribution functions. $$P(-X \leq x-a) = P(X\geq a-x) \neq 1 - P(X \leq a-x) = 1-F(a-x)$$ unless $P(X=a-x)=0$, i.e. $F(\cdot)$ is continuous at $a-x$. If you do the calculation correctly, you do end up with what the OP said, modulo typographical errors. $\endgroup$ – Dilip Sarwate Feb 26 '17 at 14:27
  • $\begingroup$ my question is wrong i dont want answer in terms of distribution function sorry about the vague title .,.,.,.,., just answer in yes or no whether converse is true or not $\endgroup$ – ANUJ NAIN Feb 26 '17 at 16:05
  • $\begingroup$ I will edit! that way. $\endgroup$ – kjetil b halvorsen Feb 26 '17 at 16:13
  • $\begingroup$ How would you define symmetric random vector? $\endgroup$ – Lisa Jan 3 '18 at 1:44
  • $\begingroup$ @kjetilbhalvorsen howcome $f (a+x) = f (a-x) $ ?$ $ let X be $N( a , 1 )$ then x-a is symmetric , then according to you $f (a+x) = f (a-x) $ , which is not true , since a +x is $ N(2a , 1) $ and a-x is $N(0 , 1 ) $ insted the correct relation should be $f (x-a) = f (a-x) $ $\endgroup$ – ANUJ NAIN Jun 1 '19 at 10:22

Just check the skewness of the distribution. If it is 0, then the distribution will be symmetric around expected value of that random variable.

Checking with probability is not a good idea. In ideal cases, it may be equal but it totally depends over the distribution. In case of normal distribution probability checking may be correct, but it all depends.

Skewness is the only way to check symmetry.

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    $\begingroup$ I dont think this answer is correct. The "skewness" (usually defined in therms of third moment) can be zero for a random variable which is not symmetric. $\endgroup$ – kjetil b halvorsen Feb 26 '17 at 11:51
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    $\begingroup$ I agree with kjetil. This is a shoot-from-the-hip response in which almost every sentence is incorrect. $\endgroup$ – Dilip Sarwate Feb 26 '17 at 14:16
  • $\begingroup$ i am confused with these answer .,.,.,.,.guys i just want to know is the converse true (as mentioned in the question )? yes or no thats it $\endgroup$ – ANUJ NAIN Feb 26 '17 at 15:13
  • $\begingroup$ I don't understand, how come here all my statements are incorrect. Also how P(x≤a−x)=P(x≥a+x), will define the symmetry in case of Poisson, Exponential, Gamma, Beta, Weibull i.e (non - normal / uniform) regression? $\endgroup$ – Neeraj Tiwary Feb 28 '17 at 2:49

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