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Suppose $X$ has a Gamma distribution $\Gamma(a,b)$. What is the distribution of $X+c$? where $c$ is a positive constant.

EDIT: I know how extract the expression of the pdf and cdf. What I wanted to know is if the resulting distribution is a know one.

Nobody
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    It might be called a shifted Gamma distribution. – Henry Feb 17 '21 at 20:18
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    If $X$ has density function $f_X(x)$ and cumulative distribution function $F_X(x)$ then $Y=X+c$ has density function $f_Y(y)=f_X(y-c)$ and cumulative distribution function $F_Y(y)=F_X(y-c)$ – Henry Feb 17 '21 at 20:19
  • I don't understand how this question without any personal work can be upvoted and answered whereas sometimes downvotes accumulate on questions showing reflection... but estimated too naive maybe... – Jean Marie Feb 17 '21 at 20:41
  • @JeanMarie, I know how extract the expression of the pdf and cdf. What I wanted to know is if the resulting distribution is a know one. – Nobody Feb 17 '21 at 20:43
  • Then say it in this way in your question (I see you have added it) – Jean Marie Feb 17 '21 at 20:44

2 Answers2

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\begin{align} & \Pr(X+c\le x) = \Pr(X\le x-c) \\[8pt] = {} & \int_0^{x-c} f(u)\, du \text{ if } x-c\ge0 \\[8pt] = {} & \int_c^x f(v-c)\, dv \\[8pt] & \text{where} \\ & v = u+c \\ & dv = dx \\ \text{so that } & \text{when } u=x-c \text{ then } v= c \\ \text{and } & \text{when } u=0 \text{ then } v=c. \end{align} So the probability density of $X+c$ is $$ g(x) = \begin{cases} f(x-c) & \text{for } x>c, \\ 0 & \text{for }x<c. \end{cases} $$

Michael Hardy
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Simply

$$f_Y(y)=\frac{b^a}{\Gamma(a)}(y-c)^{a-1}e^{-(y-c)b}$$

$y\in[c;+\infty)$

tommik
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