For $$f(x|\theta)=\frac{1}{24}x^4\theta^{-5}\exp^{-\frac{x}{\theta}}I_{(0,\infty)}(x)\hspace{0.5cm}\theta\in\mathbb{R}^+$$

I want to find a $100(1-\alpha)\%$ confidence interval.

I know $T(X_1,\dots,X_n)=\sum\limits_{i=1}^nX_i\sim\text{Gamma}(5n,\theta)$ and T is sufficient

So $$f_T(t)=\frac{t^{5n-1}e^{-t/\theta}}{\Gamma(5n)\theta^{5n}}$$ Then I used $t=y\theta$ so

$$f_Y(y)=\frac{y^{5n-1}e^{-y}}{\Gamma(5n-1)}$$ Then I should find $\gamma_1,\gamma_2$ such that $$P(\gamma_1<T/\theta<\gamma_2)=1-\alpha $$ So I need $F_Y(\gamma_1)$ and $F_Y(\gamma_2)$ and solve for $\gamma_1$ and $\gamma_2$ but I can't do it.

  • 1
    $\begingroup$ Add the self-study tag if this is homework. $\endgroup$ – StubbornAtom May 23 '20 at 18:36

You have a random sample of $n$ observations $X_1, X_2, \dots, X_n$ from a population with distribution $\mathsf{Gam}(\mathrm{shape}=5, \mathrm{scale}=\theta),$ where $\theta$ is unknown.

Suppose you wish to use these data to make a 95% confidence interval for $\theta.$ Let $T = \sum_{i=1}^n X_i,$ which has $E(T) = 5n\theta.$

You propose using the pivotal quantity $Y = T/\theta,$ which has distribution $\mathsf{Gam}(\mathrm{shape}=5n, \mathrm{scale}=1)$ and $E(Y) = 5n.$ [Note: Your density function for the distribution of $Y$ needs to have denominator $\Gamma(5n),$ not $\Gamma(5n-1).]$

If $L$ and $U$ cut probability 0.025 from the lower and upper tails, respectively, of $\mathsf{Gam}(5n,1),$ then with $n = 30,$ we can use R statistical software to find that $L=126.9562, U= 174.9372.$ [Note: In R, the second parameter is the rate $\lambda$, which is the reciprocal of the scale. But here both rate and scale are $1.]$

a = 5; n = 30
LU = qgamma(c(.025,.975), a*n, 1);  LU
[1] 126.9562 174.9372


$$ 0.95 = P(L < Y = T/\theta < U) = P(T/U < \theta < T/L),$$

so that a 95% confidence interval of $\theta$ is $(T/U,\, T/L).$

Example: As an example, we let $\theta = 9$ and use R to get a random sample of size $n = 30$ from $\mathsf{Gam}(\mathrm{shape} = \alpha = 5, \mathrm{scale} = 9).$

a = 5;  th = 9
x = rgamma(n, a, 1/th)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  11.92   28.20   45.19   46.48   58.94   99.61 
t = sum(x);  t
[1] 1394.351

The resulting 95% CI for $\theta$ is $(7.97, 10.98).$

t/qgamma(c(.975,.025), a*n, 1)
[1]  7.970578 10.982932

So my random sample was among the 'lucky' 95% of samples that produces a CI covering $\theta = 8.$

Note: If you're allergic to software, you can express the distribution $\mathsf{Gam}(5n, 1)$ as an equivalent chi-squared distribution and use printed chi-squared tables to find lower and upper boundaries.

qchisq(c(.025,.975), 300)/2
[1] 126.9562 174.9372
qgamma(c(.025,.975), 150, 1)
[1] 126.9562 174.9372
  • $\begingroup$ Thank you for your answer and how can I can express the distribution Gam(5n,1) as an equivalent chi-squared distribution? $\endgroup$ – Jhon Knows May 23 '20 at 18:20
  • $\begingroup$ I will leave that for you. If it isn't clear from your text, you can look at Wikipedia articles on gamma and chi-squared under Related distributions. (Erlang is a Gamma with integer shape parameter.) ) $\endgroup$ – BruceET May 23 '20 at 18:36

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