Chebyshev's Inequality

In this video we are going to prove Chebyshev's Inequality which is a useful inequality to know in statistics, and in the next video we are going to use it to help us prove the law of large numbers. So what is Chebyshev's Inequality? Well, it says that the probability that some random variable x differs from its mean mu by some sort of amount which is great than or equal to 'a' which is just a constant is going to be less than or equal to the variance of that random variable x divided by 'a' squared. Ok, so how do we go ahead and proof this? Well first of all we write down Markov's Inequality which is - if you remember from last time - the probability that some random variable x is greater than or equal to a is going to be less than or equal to the expected value of x divided by 'a'. So, that's just what we proved last time, but there's nothing to stop us replacing x in the above relationship with the modulus of x minus mu. That itself is a random variable, so we can just replace x in the above relationship by the modulus of x minus mu. Because this above relationship - Markov's Inequality - should hold for all random variables. So, if I do that we get that the probability of the modulus of x minus mu being greater than or equal to 'a' is less than or equal to the expected value of now not x, but its modulus of x minus mu, all divided by 'a'. So, this is starting to look a bit similar to what we have above, but the problem is the right hand side looks a bit different. We can make it quite similar though, if we realise that we can write the probability of x minus mu being greater than or equal to 'a' is actually identical to the probability that x minus mu all squared is greater than or equal to 'a-squared'. Right, so let's just think about that for a second... ...if x is less than mu, that's going to be a negative number, which, when I square it better be greater than or equal to 'a squared'. And if I was just thinking in these terms, if the difference between x and mu is negative but the absolute value of that is greater than or equal to 'a', then when I square that the absolute value of that is going to be greater than or equal to 'a squared'. And obviously, that's going to hold true if x is greater than mu, that's the trivial case. So these two probabilities are absolutely identical, but by writing the left hand side like this, we can actually replace the right hand side, just using Markov's Inequality, which is the probability that x is greater than or equal to 'a' has got to be less than or equal to the expected value of x divided by 'a'. But now we don't have x, we have on the right hand side a value of x minus mu, all squared. Right? And now, instead of 'a', we have 'a-squared'. But, in fact this is - the numerator of this function is - what the definition of the variance is. The expected value of x minus mu all squared is how we define the variance. So the top is just the variance of x, and the bottom just stays as 'a-squared'. And remembering that these probabilities up here and down here are all exactly the same, we've actually gone ahead and proved the statement which we have up here in the parenthesis. So, we've gone ahead and proved Chebyshev's Inequality. And in the next video we are going to use this to help us prove the law of large numbers. I'll see you then.