How would one derive the density or pmf of a distribution that is shifted to the right ?
For example, a Poisson distribution that is truncated at zero , is shifted to the right . And it only takes values from $x = 1$, to $x = \infty$.
How can I derive its probability mass function ? Do I derive this from the poisson distribution ? I would like to see how it is done mathematically.
1. Shifting and truncation are different things
Truncating a Poisson at zero is not the same as shifting a Poisson to the right. Shifting just moves the whole thing left or right.
Truncation actually cuts off some of the distribution. As a result the remainder is "scaled up" so that the total probability is still 1.
Here's an illustration of the difference between shifting and truncation for normal distributions. The blue is the original in both cases, the gray is the result of either shifting or truncation:
2. Writing down a shifted probability function
Actual shifting (by $\delta$, say) is easy to write down, you just "replace" the variable ($x$, say) by $(x-\delta)$ (and the domain is shifted by $+\delta$). Slightly more formally:
So consider a random variable $X\sim\text{Poisson}(\lambda)$
$$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!},\quad x=0,1,2,\ldots$$
Now consider $Y=X+1$; $Y$ is just $X$ shifted 1 to the right.
$$P(Y=y)=\frac{e^{-\lambda}\lambda^{y-1}}{(y-1)!},\quad y=1,2,\ldots$$
(The required "0 elsewhere"'s being understood as needed.)
3. Writing down a truncated probability function
Since you seem to be looking at left-truncation, I'll discuss that specifically, but right-truncation or truncation at both ends works analogously.
Imagine your probability function or density function is $f_X(x),\quad l \leq x\leq u$ but now you have a new variable, $Y$, which is distributed like $X$ but truncated on the left at $t>l$ (i.e. that any values $\leq t$ are not observed*). Then $f_Y(y) = \frac{1}{1-F_X(t)} f_X(y),\quad t< y\leq u$.
* If truncation at $t$ instead means values $<t$ are not observed it changes to $\frac{1}{1-F_X(t^-)} f_X(y),$ $t\leq y\leq u$ (where $(t^-)$ is understood in the same sense as $(a-)$ here).
Again, consider a Poisson random variable $X\sim\text{Poisson}(\lambda)$
$$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!},\quad x=0,1,2,\ldots$$
Let $Y$ be like $X$ but truncated at $0$ (which here can only mean that $0$ is truncated; the alternative meaning does nothing). Then
$$P(Y=y)=\frac{e^{-\lambda}\lambda^y}{y!(1-e^{-\lambda})},\quad y=1,2,\ldots$$
We can see how these look different by comparing (for a Poisson with $\lambda=1.8$) shifting up by 1 vs truncating 0:
Additional discussion of truncation can be found here at Wikipedia.
If $X$ has a given distribution $f(x)$, then the distribution of $X+a$ is what you call the shifted distribution and is $g(x) = f(x-a)$. For discrete distributions, all the probability masses shift to the right by $a$ as in Glen_b's answer.
On the other hand, the truncated distribution is effectively the conditional distribution of $X$ conditioned on the event that $X >b$ or $X \geq b$ (or, if you lop off the top, conditioned on $X < c$ or $X \leq c$). For the case of the conditioning event being $\{X > b\}$, this distribution is given by $$h(x) = \begin{cases}\frac{f(x)}{P\{X > b\}}, & x > b,\\0, & x \leq b, \end{cases}$$ and similarly for the other possible conditioning events.
Yet another possibility for truncation is that we have a new random variable $Y$ that is related to $X$ as $$Y = \begin{cases}X, & X > b,\\b, & X \leq b,\end{cases}$$ in which case, $Y$ has a mixed distribution if $X$ is a continuous random variable, with a distribution that has a point mass at $b$ and is continuous to the right of $b$. But, for a discrete random variable $X$, $Y$ is also a discrete random variable and its probability mass function is given by $$p_Y(x) = \begin{cases}p_X(x), & x > b,\\P\{X \leq b\}, & x = b, \\0, & x < b.\end{cases}$$
