I am struggling with finding a method of moments estimator for (seemingly) simple situation: pdf is given by $P_\theta(X = x) = \frac{1}{\theta}$, $x \in$ {1,2,...$\theta$}, where $\theta \in N$.

My idea was to find the expectation and then, the first sample moment. I did so far:

$E(x) = \int_{0}^{\theta}xf(x)dx$

$E(x) = \int_{0}^{\theta} \frac{1}{\theta}xdx$

$E(x) = \frac{1}{\theta}\int_{0}^{\theta}xdx$

$E(x) = \frac{1}{\theta}\int_{0}^{\theta}xdx$


Applying the limits, I get:

$\frac{1}{\theta}(\frac{\theta^2}{2} - \frac{1}{2})$,

from which follow that:

$\bar{X} = \frac{\theta}{2} - \frac{1}{2\theta}$.

This does not seem like a correct answer, because the book provides MOM as:

$2\bar{X} -1$

But I cannot get the same MOM with the expectation I calculated. I admit that my expectation is wrong, or I misunderstand the further steps of calculation.

I would appreciate if someone can help!

Thank you!



This is a discrete distribution, so it does not have a density $f(x)$ but instead a probability mass function. Its expectation is $$E[X]= 1 \times \frac1\theta + 2 \times \frac1\theta + \cdots + \theta \times \frac1\theta = \frac{\theta(\theta+1)}{2} \times \frac1\theta = \frac{\theta+1}{2}$$

which should not be a surprise as it is the middle of a uniform distribution from $1$ through to $\theta$.

Solving this for $\theta$ gives $\theta = 2 E[X] -1$ making the method of moments estimator $\hat \theta = 2 \bar X -1$

  • $\begingroup$ Thank you, Henry! You clarified perfectly, I really missed out that the distribution was discrete. $\endgroup$
    – marina_esp
    Jan 17 '20 at 14:43

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