Will adding a constant to a random variable change its distribution? Suppose I have a random variable $X$ and to this, I add a constant $c>0$. Will $X+c$ have a different distribution? From my intuition it seems so, but I am unable to prove it from a measure-theoretic point of view. Does the proof require measure theory? Thanks!
 A: I feel like I'm making a huge mistake because of how simple this feels, but...
If $X$ and $X+c$ had the same law, they would have the same expectation. But alas, $\Bbb E [X + c] = \Bbb E X + c$ so unless $c=0$, they have different laws.
A: Let $c\gt 0$. We will show that $X$ and $X+c$ never have the same distribution, because they have different cumulative distribution functions.  We will show this by showing that there is a positive $h\lt c$ and a number $x$, such that $\Pr(X\le x+h)\ne \Pr(X\le x)$, and therefore $\Pr(X\le x-c)\ne \Pr(X\le x)$.
For if $F(t)$ is the cumulative distribution function of $X$, then $\lim_{t\to-\infty} F(t)=0$ and $\lim_{t\to\infty} F(t)=1$. So there is an $a$ such that $F(a)\lt 1/4$ and a $b$ such that $F(b)\gt 3/4$. By stepping forward from $a$ by steps of length $h$, we can get beyond $b$ in a finite number of steps. If we always had $F(t+h)=F(t)$, we would have $F(c)=F(a)\lt 1/4$ for some $c\gt b$, which is impossible.
A: Let $Y=aX+b$; a linear transformation.   The distribution of the random variable $Y$ can then be said to have the same shape as that of $X$, but with a change of scale and a shift of ordinates.   Thus while it clearly is not the same distribution (unless $a=1 \wedge b=0$), it is the same family of distribution.
Eg: $X_1\sim \mathcal U(c; d)$ then $aX_1+b \sim \mathcal U(ac+b; ad+b)$
Eg: $X_2\sim\mathcal N(\mu, \sigma^2)$ then $aX_2+b \sim\mathcal N(a\mu+b, a^2\sigma^2)$
Etc…
A: Suppose $X\sim N(0,1)$.  Then $\Pr(-1<X<1) \approx 0.68$ but $\Pr(-1<X+3<1) <0.03$.
So $X$ and $X+3$ have different distributions.
