Given the following uniform distributions \(d_1\), \(d_2\), \(d_3\), what can you say about their mean \((\mu_i)\)?
The given distributions are uniform. A random variable \(X\) has uniform distribution if it’s pdf: \[f(x) = \begin{cases} \frac {1}{b-a} & \text{for x $\in$ [a,b]}\\ 0 & \text{otherwise}\\ \end{cases} \] That is, pdf of \(X\) is constant in the range [a,b]. The mean of \(X\) is: \[\begin{aligned} E[X] &= \int xf(x)dx \\ &= \int_{a}^{b} \frac {1}{b-a}xdx + 0 \\ &= \frac {b-a}{2}\\ \end{aligned} \] So the mean of \(X\) is the mid point \([a,b]\) i.e. we just need to find the midpoint of the values of x where it’s pdf is not zero.
So for the above given graph, \(\mu_1 < \mu_2 < \mu_3\).
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