Sampling from product of exponential distributions

Question

I have a distribution who's moment generating function is the product of two exponentially distributed variables moment generating functions. If I wanted to generate samples from the distribution, would it be appropriate to generate a random sample from the exponential distribution and multiply it by two? For clarification the PDF is below:

$$f(x) = \lambda^2 x\cdot exp(-\lambda x), x \geq 0 $$

The MGF calculated for this PDF was:

$$M(t) = \lambda^2/(\lambda - t)^2$$

Answer 1

The exponential distribution has the scaling property. If $X \sim exp(\lambda)$, then $k \cdot X \sim exp(\lambda/k)$ for any $k > 0$, in particular $k = 2$, and the pdf of $exp(\lambda/2)$ is not equal to the expression you are giving, so the answer to your question is no.

Answer 2

The moment generating function of the sum of two independent random variables is the product of their individual moment generating functions: $$E\left[e^{(X+Y)t}\right] =E\left[e^{Xt}e^{Yt}\right] =E\left[e^{Xt}\right]E\left[e^{Yt}\right]$$

so just generate two samples from an exponential random variable with parameter $\lambda$ and add them together

This is not the same as generating one sample and multiplying it by $2$

Answer 3

Exponential distribution. If $X \sim \mathsf{Exp}(rate = \lambda),$ then its PDF is $f_X(x) = \lambda e^{-\lambda t},$ for $t,\lambda > 0.$ Its MGF is $m_X(t) = \frac{\lambda}{\lambda - t},$ for $t > 0.$ The CDF is $F_X(x) = 1 - e^{-\lambda t},$ for $t > 0.$

Gamma distribution. If $X_1$ and $X_2$ are independent random variables distributed as $\mathsf{Exp}(rate = \lambda),$ then $Y = X_1 + X_2 \sim \mathsf{Gamma}(shape = 1, rate = \lambda),$ with PDF $f_Y(y) = \lambda^2 y e^{-\lambda t},$ for $y, \lambda > 0.$ The MGF is the square of the exponential MFG $m_Y(t) = (\frac{\lambda}{\lambda - t})^2,$ for $t > 0.$

The quantile method of generating an exponential random variable. If $U^\prime \sim \mathsf{Unif}(0,1),$ then setting $U = F_X(X)$ and solving for $X$ in terms of $U^\prime,$ we have $X = -\log(1-U^\prime)/\lambda.$

Most pseudorandom number generators produce output that is (for practical purposes) indistinguishable from $U \sim \mathsf{Uniform}(0,1).$ So we can generate $X \sim \mathsf{Exp}(\lambda)$ as $X = -\log(U)/\lambda,$ because if $U^\prime \sim \mathsf{Unif}(0,1),$ then also $U = 1 - U^\prime \sim \mathsf{Unif}(0,1).$ This is called the 'inverse CDF' or 'quantile' method of generating an exponential random variable.

Generating gamma random variables. Then finally, you can generate a sample of size $n=2$ from an exponential distribution with rate $\lambda = .2,$ and then add them to generate a gamma random variable, according to the following statement in R statistical software:

u = runif(2);  y = sum(-log(u)/.2)

The object u is a vector of realizations of two standard uniform random variables. The object y is a sample of size 1 from a gamma distribution. Exponential and gamma distributions are programmed into R, so y could be generated as y = rgamma(1, 2, .2), but that shortcut wouldn't illustrate the method of your Question.

Simulating a gamma distribution. Replicating this $m = 10,000$ times, we can simulate the a random sample of $m$ random variables $Y \sim \mathsf{Gamma}(2, \lambda):$

n = 2;  m = 10^4;  lam = .2
y = replicate(m, sum(-log(runif(n))/lam)) 
hdr = "Simulated Dist'n of GAMMA(2, .2) with PDF"
hist(y, br=40, prob=T, col="skyblue2", main=hdr)
  curve(lam^2*x*exp(-lam*x), 0, 80, add=T, lwd=2, col="blue")
  abline(h=0, col="green3"); abline(v=0, col="green3")

Note: Most of the information about exponential and gamma PDF, MGF, and CDF (and some of the information about generating a gamma sample) can be found in Wikipedia articles about those distributions. Because you asked about generating gamma random variables, I suppose information on that is also contained in your textbook or class notes.