obtaining the true values for two parameter gamma distribution please i have a data set on length of stay in the hospital for two parameter gamma distribution. i want to compare moment and maximum likelihood estimation method for the distribution using mse. i was able to obtain moment estimate for alpha and beta and i was also able to obtain maximum likelihood estimate for alpha and beta. Now my question is that how do i know the true values so that i would be able to compare the two estimators. thank you in advance.
 A: Here is a brief outline of the rationale and R code for getting parametric bootstrap 95% confidence
intervals for the unknown shape parameter $\alpha$ and scale parameter $\beta$ of
a gamma distribution based on a random sample of $n$ observations.
Let $X_1, \dots X_{269} \stackrel{iid}{\sim}\mathsf{Gamma}(\alpha, \lambda =
1/\beta),$ where $\lambda = 1/\beta$ is the rate parameter (used in R).
We seek a 95% CI for $\beta.$ The observed MME is 
$\tilde \beta_{obs} = \frac{n-1}{n}\frac{S^2}{\bar X}.$
If we knew the distribution of $V_\beta = \tilde \beta/\beta,$ we could find
constants $L_\beta$ and $U_\beta$ such that
$$.95 = P(L_\beta < V_\beta < U_\beta) = 
P\left(\frac{\tilde\beta_{obs}}{U_\beta} < \beta <\frac{\tilde\beta_{obs}}{L_\beta}\right),$$
and $(\tilde\beta_{obs}/L_\beta,\, \tilde\beta_{obs}/L_\beta)$ would be a
95% CI for $\beta.$ (Note the reversal of inequalities because of taking
reciprocals to isolate $\beta.$) 
Not knowing the distribution of $V_\beta,$ we use a parametric bootstrap to 
obtain estimates $L_\beta^*$ of $L_\beta$ and $U_\beta^*$ of $U_\beta$.
Similarly for $\alpha,$ estimated by $\tilde\alpha = \bar X/\tilde\beta.$

Entering the so-called bootstrap world, we take a large number $B$ of
  're-samples' of size $n$ from 
  $\mathsf{Beta}(\tilde\alpha_{obs}, 1/\tilde\beta_{obs}),$
   temporarily using the observed estimates as proxies for the unknown parameters.
  For each of the $B$ re-samples, we find $\tilde \alpha^*$ and $\tilde\beta^*$,
  and from them $V_\beta^* = \tilde\beta^*/\tilde\beta_{obs}$ and
  $V_\alpha^* = \tilde\alpha^*/\tilde\alpha_{obs}.$

Returning to the 'real world', we take quantiles .025 and .975 of the many $V_\beta^*$ to get the desired
estimates $L_\beta^*$ and $U_\beta^*$, and similarly for $L_\alpha^*$ and $U_\alpha^*.$ Thus we have parametric bootstrap 95% CIs
$(\tilde\beta_{obs}/U_\beta^*,\, \tilde\beta_{obs}/L_\beta^*)$ for $\beta$ and 
$(\tilde\alpha_{obs}/U_\alpha^*,\, \tilde\alpha_{obs}/L_\alpha^*)$ for $\alpha.$

In the R code below, we use .re for $*$, sc for $\beta$ sh for $\alpha,$
and a for $\bar X.$
set.seed(1212)
B = 10^5;  vsh.re = vsc.re = numeric(m)
n = 269;  a = 7.78;  s = 8.4
sc.obs = (n-1)*s^2/(n*a);  sh.obs = a/sc.obs
for(i in 1:B)  {
  x.re = rgamma(n, sh.obs, 1/sc.obs)
  a.re = mean(x.re);  s.re = sd(x.re)
  sc.re = (n-1)*s.re^2/(n*a.re);  sh.re = a.re/sc.re 
  vsc.re[i] = sc.re/sc.obs;  vsh.re[i] = sh.re/sh.obs  }
UL.vsc = quantile(vsc.re, c(.975,.025));  CI.sc = sc.obs/UL.vsc
UL.vsh = quantile(vsh.re, c(.975,.025));  CI.sh = sh.obs/UL.vsh
CI.sc; CI.sh
##     97.5%      2.5% 
##  6.916052 11.989209   # CI for scale 'beta' 
##     97.5%      2.5% 
##  0.6727584 1.0926581  # CI for shape 'alpha'
sc.obs;  sh.obs
## 9.035693              # Observed scale
## 0.8610297             # Observed shape


Notes: (1) Now that I have seen your estimates, I'm wondering if the true distribution
might be exponential instead of gamma. The shape parameter $\alpha = 1$ would
be consistent with exponential data. If you used a distribution ID procedure,
then of course it will get a better 'fit' using two parameters, shape and
scale, rather than one. When using distribution ID procedures, it is
always difficult to decide whether the most realistic fit is to the
family or a subfamily with only marginally worse fit.
(2) A bootstrap procedure for MLEs would require a numerical solution
for the estimate of the shape parameter to be done afresh on each passage through the loop.
(3) For bootstrapping a location parameter $\tau,$ one would re-sample to
obtain information about $D = \tilde\tau - \tau$ rather than $V = \tilde\tau/\tau.$
