I'm having some issues working through this practice problem. I have worked through the first portion of it, and I have the solution, but I don't understand how/why the solution does two things at the end.


Use the rejection method to sample from the gamma density $\Gamma(\lambda, t)$ where $t (\geq 1)$ may not be assumed integral. [Hint: You might want to start with an exponential random variable with parameter $\frac{1}{t}$.]

What I've done

Let $Z\sim \Gamma(1, t)$. This can be done without loss of generality because, for such a Z, $\frac{Z}{\lambda}\sim \Gamma(\lambda, t)$.

Let $V\sim Uniform(0,1)$. Using the hint, for some $y\in (0,1)$,

\begin{align*} F(x) &= 1-e^{\frac{-x}{t}}\\ y &= 1-e^{\frac{-x}{t}}\\ x &= F^{-1}(y) = -t ln(1-y)\\ X &= -t~ln(1-V)\\ X &= -t~ln(V) \end{align*}

(Since if $1-V\sim Uniform(0,1)$, then $V\sim Uniform(0,1)$.)

Now, we need to find some $c$ such that

\begin{align*} f_Z(x) \leq c f_X(x) \end{align*}

where $f_Z(x) = \frac{1}{\Gamma(t)}x^{t-1}e^{-x}$ and $f_X(x)= \frac{1}{t} e^{\frac{-x}{t}}$.

Thus, we need to find a $c$ that satisfies:

\begin{align*} \frac{f_Z(x)}{f_X(x)} &\leq c\\ \frac{\frac{1}{\Gamma(t)}x^{t-1}e^{-x}}{\frac{1}{t} e^{\frac{-x}{t}}} &\leq c\\ \frac{t e^{x(1/(t-1))}x^{t-1}}{\Gamma(t)} &\leq c \end{align*}

By taking the derivative w.r.t. X, I verified that, for $t\geq 1$ and $x>0$, the LHS is maximized by $x=t$ (I'll omit this work, but I found that the derivative had a root at $x=-t^2+2t-1$, so over the ranges of $t$ and $x$, $t>x$, thus the LHS is strictly smaller than its value when $x=t$).

Thus, the smallest value of $c$ such that $LHS\leq RHS$ is:

\begin{align*} c &= \frac{1}{\Gamma(t)}t^{t} e^{-(t-1)} \end{align*}

What I don't understand

The solution uses the following two things to continue the problem that I don't understand.


"Conditional on the event $A$:

\begin{align*} U &\leq \frac{X^{t-1}e^{-t}}{\Gamma(t)}t~e^{\frac{-X}{t}} \end{align*}

where $U\sim Uniform(0,1)$, X has the required gamma distribution."

I don't understand where this definition of $A$ comes from.

I can see that $\frac{X^{t-1}e^{-t}}{\Gamma(t)}$ is almost $f_Z(x)$, except that $e^{-x}$ has been replaced by $e^{-t}$ and the rest is almost $f_X(x)$, except $\frac{1}{t}$ has been replaced with $t$.

But, even if that were the case, $A$ would be showing that $U\leq$ the joint density of the two component functions (and, again, I know that this is not the case, but it is the closest explanation I can come up with), not the $\frac{f_Z(x)}{c~f_X(x)}$ that I believe I need to continue the problem. I cannot manipulate $\frac{f_Z(x)}{c~f_X(x)}$ to create this $A$.

I understand that the next step likely involves the fact that $F_Z(z;1, t)$ will take the form $\frac{\gamma(t, z)}{\Gamma(t)}$, and $\gamma(t, z) = \int_0^z x^{t-1}e^{-x} dx$. I just don't know what to do with this information, given that I don't know where this $A$ came from or how to use it.


"We note that $A = \{log(U)\leq(n-1)(log(X/n)-(X/n)+1) \}$."

Where did this come from and why is it worth noting?

  • 1
    $\begingroup$ This is a special case of the question answered very nicely at stats.stackexchange.com/a/41284/919. $\endgroup$
    – whuber
    Nov 3, 2019 at 22:36
  • $\begingroup$ That helps my intuition somewhat, but I'm not sure it really answers my question. I feel confident that I have already identified the relevant envelope function, at least insofar as it satisfies $f_Z(x) \leq c f_X(x)$, which is how I currently understand envelope function working in this context. If that understanding is in error, I would appreciate some intuition as to why finding $c$ is necessary but not sufficient to define event $A$ in this context. (cont'd) $\endgroup$
    – mdawgig
    Nov 4, 2019 at 1:37
  • $\begingroup$ I want to understand why that particular value of $A$ was chosen in the solution I have, given that it does not align with any way that I can manipulate the RHS of $\mathbb{P}(U\leq \frac{f_Z(x)}{c~f_X(x)})$. Upon further thought, I feel confident that my goal is to create an ultimate ratio of integrals such that I have: $\frac{\frac{1}{\Gamma(t)}\frac{1}{c} \int_0^z x^{t-1}e^{-x} dx}{\frac{1}{c}}$ (I have omitted the integral of the denominator because it integrates to 1.) (cont'd) $\endgroup$
    – mdawgig
    Nov 4, 2019 at 1:37
  • $\begingroup$ This will give me $\frac{\frac{\gamma(t, z)}{\Gamma(t)} \frac{1}{c}}{\frac{1}{c}} = F_Z(z)$ with mean number of attempts $c$ before a sample of $n=1$ is obtained. I just feel like I am missing one crucial step that gets me from my definition of $c$ to the answer's definition of $A$, and if your link provides that answer, I'm missing some other piece of information that connects the two. $\endgroup$
    – mdawgig
    Nov 4, 2019 at 1:37

1 Answer 1


The rejection method can be explained either by an envelope argument, namely that pairs $(X,U)$ that are generated uniformly on the set $$\mathfrak S_{cf_X}=\{(x,u);\ 0\le u\le c f_X(x)\}$$ [called the subgraph of $cf_X$] are such that

  1. $X\sim f_X(x)$ (we call this result the fundamental lemma of simulation in our book);
  2. if $U<f_Z(X)$ then $X\sim f_Z(x)$

This can be seen on the picture below where the hollow circles are $(X,U)$'s that are generated uniformly over a square that contains the graph of the target $f_Z$ (a Gamma truncated to (0,1) in this illustration) and the filled circles are those that fall below $f_Z$. They are indeed uniformly distributed over $\mathfrak S_{f_Z}$.

enter image description here

It can also be explained by a stopping rule reasoning: if one runs a sequence of iid simulations from $f_X$, $X_1,X_2,\ldots$ and an associated auxiliary sequence of uniforms $U_1,U_2,\ldots$ until $$U_i<f_Z(X_i)/cf_X(X_i)$$ happening for the (random) index value $N$, then $$X_N \sim f_Z(x)$$ since the density of $X_N$ is proportional to$$f_X(x)\times\mathbb P(U\le f_Z(X)/cf_X(X)\mid X=x) \propto f_Z(x)$$

Note: I do not understand the part of the question that mentions "The solution" as I cannot tell where this solution is coming from.


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