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I don't think that's a good representation of what they're saying. They showed that the average number of friends that your friends will have is mu+std/mu, so i
by xypage 4y ago
I don't think that's a good representation of what they're saying. They showed that the average number of friends that your friends will have is mu+std/mu, so if the distribution is anything but perfectly homogeneous then you can expect that your friends will have more friends than you. It doesn't need to be super skewed with some people knowing everybody, any variation _at all_ leads to this, no need for outliers.
- pessimizer 4y ago> any variation _at all_ leads to this, no need for outliers. A lot of dismissals miss this in talk about super popular celebrities. It just requires that some people have more friends than other people. The people who have more friends are more likely to be friends with you than the people who have fewer friends [edit: even if "more friends" means five friends and "fewer friends" means four friends.] That's it. It isn't about outliers pulling up means.
- WastingMyTime89 4y agoTheir description seems like a fairly good take. The main idea is that in graph you are more likely to be connected to an outlier when it comes to connexions by virtue of them being more connected and these outliers skew the average. Of course you might yourself be the outlier. If you think about it, this is not far from saying that: if you know a distribution is non homogenous with a limited number of elements skewed - like say a normal distribution - and pick an element at random, it is more likely to be average than exceptional which is pretty trivial when you think about it.