Show that the random variable $V=X-a$ and $U=a-X$ have same distribution?

Given $X$ is a continuous random variable whose density is symmetric about a point $a$.

Show that $V=X-a$ and $U=a-X$ have same distribution.

$$F_U(u) = P(U \leq u) = P(X-a \leq u) = F_X(a+u)$$ and similarly $$F_W(w) = 1 - F_X(a-w) \longrightarrow f_U(w) = f_X(a+w) = f_X(a-w)$$ by symmetry. Therefore, $f_U(u) = f_X(a+u)$ by changing variable $w$ to $u$, which shows $f_U(u)=f_W(u)$. Is this solution right? Thanks!

• Can you solve the problem for the special case $a=0$? – Dilip Sarwate Sep 18 '13 at 17:57
• user30438, please emulate the markup in the first half of your question to make the second half readable. – whuber Sep 18 '13 at 19:18
• I have typeset the second half of your post to enhance readability. Please check that I haven't inadvertently changed the meaning of anything. – Macro Sep 20 '13 at 23:36
• You can do this directly from the definition of the distribution function and using the given symmetry. – Glen_b -Reinstate Monica Sep 21 '13 at 2:33

1 Answer

• It is sufficient to show that $U$ and $V$ have the same cumulative distribution function. To see that, notice that $\mathbb P(a-X\leqslant t)=\mathbb P(a-t\leqslant X)=\int_{a-t}^\infty f_X(s)\mathrm ds$, then use a substitution. (notice that $\mathbb P\{a-t=X\}=0$ since $X$ has a density)

• Express the $(2p+1)$-th moment in terms of an integral involving $f_X$, then use a substitution.

• You're right (+1), but why go through the integration when you really need only to invoke basic properties of inequalities? To wit, the events $V\le t$, $X-a\le t$, and $X\le a+t$ are identical (these are just equivalent descriptions of the same set); $\Pr(X\le a+t) = \Pr(X\ge a-t)$ is what it means for $X$ to be symmetric about $a$; and the events $X\ge a-t$, $-X\le t-a$, $a-X\le t$, and $U\le t$ are identical. Consequently $F_V(t)=F_U(t)$ for all numbers $t$, meaning $V$ and $U$ have the same distribution. The continuity assumption was not needed anywhere. – whuber Sep 18 '13 at 18:18
• @whuber: I think Davide is trying to draw out a subtle point about the wording of the question. If it had been stated "Let $X$ be a symmetric random variable with density $f$..." there would be little doubt. The question instead simply states that "$X$ is a continuous random variable with symmetric density $f$..." That leaves something to be proved. Whether that was the intent of the question is another matter altogether. (The slight distinction may well have been lost on the one originally posing it!) – cardinal Sep 18 '13 at 20:29
• @Cardinal I agree that some mechanism to relate a density to probability is needed. But historically such things could be done long before Calculus was invented, so perhaps a simple intuitive demonstration is still possible. Were we to draw the graph of the density $f$ and note its mirror symmetry around $x=a$, then would it not be immediate that the region to the left of $a+t$ is congruent to the region to the right of $a-t$? From that, the formula for the symmetry of the CDF follows easily. (Technically, we only need invariance of Lebesgue measure under reflections in $\mathbb{R}$.) – whuber Sep 18 '13 at 21:40
• @user30438 Nope: many distributions do not have pdfs. A distribution of a random variable $X$ is defined to be its CDF. – whuber Sep 21 '13 at 1:10
• @COOL The simplest is any Bernoulli distribution: it is singular with respect to Lebesgue measure, whence it has no PDF. – whuber Sep 23 '13 at 14:50