# Likelihood function of the gamma distribution

• safina

#### safina

There is a random sample of size n from a gamma distribution, with known r. Please help me formulate the likelihood function of the gamma distribution.

I understand that the density function is the following:
$$f\left(y;r,\lambda\right)=\frac{\lambda}{\Gamma\left(r\right)}\left(\lambda x\right)^{r-1}e^{-\lambda x}$$

I also understand that the likelihood function is the product of the individual density functions.
Assuming independence, I write it as:
$$L\left(\underline{y};r, \lambda\right)=\left[f\left(y;r,\lambda\right)\right]^{n}$$
$$=\left[\frac{\lambda^{r}y^{r-1}e^{-\lambda y}}{\Gamma\left(r\right)}\right]^{n}$$

I am now stuck with the product of the $$y^{r-1}$$ and $$\Gamma\left(r\right)$$.

Please help me what to do, since I need the answer to find the maximum likelihood estimator of $$\lambda$$.

Take the log of both sides. The log function is monotonic so "$\lambda$ maximizes log L" iff "$\lambda$ maximizes L."

Take the log of both sides. The log function is monotonic so "$\lambda$ maximizes log L" iff "$\lambda$ maximizes L."

Okay, thank for that. Can you help me further for the exact form of the likelihood function so that I can take the log on both sides afterwards?

I also understand that the likelihood function is the product of the individual density functions.
Assuming independence, I write it as:
$$L\left(\underline{y};r, \lambda\right)=\left[f\left(y;r,\lambda\right)\right]^{n}$$

Not quite - the likelihood function is
$$L\left(\underline{y};r, \lambda\right)=\prod_{i=1}^n f\left(y_i;r,\lambda\right)$$
since it's for a sample of size n. After taking the log and differentiating with respect to $\lambda$ you'll find that terms like $\Gamma(r)$ disappear.

Not quite - the likelihood function is
$$L\left(\underline{y};r, \lambda\right)=\prod_{i=1}^n f\left(y_i;r,\lambda\right)$$
since it's for a sample of size n. After taking the log and differentiating with respect to $\lambda$ you'll find that terms like $\Gamma(r)$ disappear.

Alright, thank you for all your replies. I've tried figuring them out. Here are the outcomes. Kindly check if these are right.

$$\frac{d}{d\lambda} log L\left(\underline{y}; r, \lambda\right) = \frac{nr}{\lambda} - \sum y$$
Equating the derivative above to zero results to:
$$\frac{nr}{\hat{\lambda}}= \sum y$$
solving for $$\hat{\lambda}$$, I have replaced $$n\bar{y}$$ for $$\sum y$$, and were able to come up with an equation $$\hat{\lambda} = \lambda$$.

Is this the result am I suppose to have?

If this really is it, is this MLE unbiased?

You may not assume sample average = distribution mean. The sample average is just a random variable, like y itself; it does not have a constant value.

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You may not assume sample average = distribution mean. The sample average is just a random variable, like y itself; it does not have a constant value.

Oh, here's what I've done.
$$\frac{nr}{\hat{\lambda}}= \sum x$$
Solving for $$\lambda$$:
$$\hat{\lambda} = \frac{rn}{\sum x} = \frac{rn}{n \bar{x}} = \frac{rn}{n \frac{r}{\lambda}} = \lambda$$

Is this not right?

I haven't checked your math, but assuming that you haven't made a mistake, you should stop at lambda hat = r n / Sum(x). That's your MLE of lambda.