# Scale the gamma distribution

I have a question regarding the distribution of a random variable $$\sum_{i=1}^{n} X_{i}^{k}$$ given that we know that $$\frac{1}{\theta} \sum_{i=1}^{n} X_{i}^{k} \text { has the Gamma }(n, 1) \text { distribution. }$$

I know that the gamma distribution has the scaling property so the first parameter should be $$\mathrm {Gamma'}(n/\theta, ?)$$ where $$\mathrm {Gamma}'$$ denotes our desired distribution, yet I don't know how to find the second parameter. Any help would be much appreciated.

• The first parameter doesn't change: it's the shape parameter. What you do with the second parameter depends on whether it's a scale or rate (inverse scale)--but that description contains the answer to your question, because $\theta$ scales the Gamma distribution up to the sum of the $X_i^k.$ – whuber Oct 14 '20 at 16:10
• @whuber According to this, the first paramter does change. right?math.wm.edu/~leemis/chart/UDR/PDFs/GammaS.pdf – ceins Oct 14 '20 at 17:33
• You need to pay attention to the conventions: your reference swaps the parameters. – whuber Oct 14 '20 at 18:24
• @whuber , so, to see if its clear to me, then $\sum X_i$ should be $Gamma′(nθ,1)$ and thus have mean nθ and expectation nθ^2? This is confusing to me..Thank you – ceins Oct 14 '20 at 21:35
• A basic property of scaling is that when you multiply any number by $\theta$ -- which is what you must do to obtain $\sum X_i^k$ from a Gamma variate -- its square is multiplied by $\theta^2.$ Thus, when you multiply a random variable by $\theta$ its expectation must be multiplied by $\theta$ and its variance by $\theta^2.$ – whuber Oct 14 '20 at 22:37