I am confused about the steps I need in order to solve the equation below. I must use conditional distribution (and NOT the factorization theorem).

Q: $X_1, . . . , X_n$ is a random sample from a Gamma(2, θ) distribution.

Show that the following is a sufficient statistic for θ: $$T(\mathbf{X}) = \sum_{i=1}^n X_i$$ Essentially, I just need show that the conditional density $f(x_1, . . . , x_n|T(\mathbf{X}))$ doesn't depend on θ.

Step 1: find the pdf of the gamma function

Step 2: let $T(x)= θ$ (all θ's in $X_i$)

Step 3: the joint density i$$p(x|θ)= \prod_{i=1}^n \frac{1}{Γ(2)*θ^2}*x^{2-1}*e^{-x/θ} = \frac{1}{θ^2}\sum_iX_i*e^{-\sum_i X_i/θ}$$

Note: $Γ(2)= (2-1)!=1!=1$ (correct?)

Step 4: use conditional probability: $p(x|θ)/q(T(x)|θ)$ to prove $T(\mathbf{X})$ does not depend on θ by plugging in $p(x|θ)$ into the equation, and $q(T(\mathbf{X}|θ)= 1/θ^2 \sum_iX_i*e^{-\sum_i X_i/θ}$.

This means the answer will be one. What am I missing for $p(x|θ)$? My answer should result with the θ's cancelling out when using conditional probability, i.e. $p(x|θ)/q(T(\mathbf{X}|θ)$ will cancel out the θ's.

  • $\begingroup$ Perhaps try wrapping some of your mathematical expressions in the $ tags so that they use latex this will make your expressions a little easier to read. $\endgroup$
    – analystic
    Commented Feb 16, 2015 at 3:42

1 Answer 1


Step 2 does not make sense

Step 3 has the rhs wrong$$p(\mathbf{x}|θ)= \prod_{i=1}^n \frac{1}{Γ(2)θ^2} x_i^{2-1} e^{-x_i/θ} = \frac{1}{θ^{2n}}\prod_ix_i e^{-\sum_i x_i/θ}$$

Step 4 has the density of $T(\mathbf{X})$ wrong$$q(t|θ)=\frac{1}{\Gamma(2n)θ^{2n}}t^{2n-1}e^{-t/θ}$$

Answer The ratio $$p(\mathbf{x}|θ)/q(T(\mathbf{x})|θ)=\frac{1}{θ^{2n}}\prod_ix_i e^{-\sum_i x_i/θ}\big/\frac{1}{\Gamma(2n)θ^{2n}}T(\mathbf{x})^{2n-1}e^{-T(\mathbf{x})/θ}=T(\mathbf{x})^{-2n+1}\prod_ix_i\Gamma(2n)$$does not depend on θ.


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