Probability Bounds

What this lecture covers

This problem bounds the probability that the sum of ten independent uniform(0,1) random variables exceeds 7, using three progressively more informative techniques: the Markov inequality, the Chebyshev inequality, and the central limit theorem. It shows concretely how each method uses more information about the distribution to produce a tighter bound on the same probability.

It draws directly on the Markov and Chebyshev inequalities and the central limit theorem developed earlier in the course. After watching, you should be able to apply all three techniques to the same tail-probability problem and explain why they give progressively tighter results.

Key ideas

• Setup: X = X1 + ... + X10, each Xi uniform on [0,1], independent; the goal is bounding P(X >= 7), where E[X] = 5.
• Markov inequality bound: using only E[X] = 5, P(X >= 7) <= E[X]/7 = 5/7, a loose bound.
• Chebyshev inequality bound: rewriting the event as a two-sided deviation from the mean and using the symmetry of the sum around 5, plus the variance Var(X) = 10/12, tightens the bound to 5/48, roughly 0.10.
• Central limit theorem bound: standardizing X and treating it as approximately standard normal gives P(X >= 7) ~ 1 - Phi(2.19) ~ 0.014, far tighter than either inequality-based bound.
• More information yields tighter bounds: each method uses progressively more distributional information (mean only, then mean and variance, then an approximation to the full shape), and the bounds improve accordingly, from 0.71 to 0.10 to 0.014.
• Concentration around the mean: the shrinking bounds illustrate that the sum of even a modest number of independent random variables concentrates fairly tightly around its expected value.

Before you watch

• Review the Markov and Chebyshev inequalities and the central limit theorem from the earlier lectures in this unit.
• Be comfortable computing the mean and variance of a uniform random variable and of a sum of independent random variables.

Check your understanding

1. Why does the Chebyshev bound require rewriting P(X >= 7) as a two-sided deviation from the mean before it can be applied?
2. Why does using the central limit theorem produce a tighter bound than Chebyshev's inequality in this example?
3. What does the progression from 5/7 to 5/48 to about 0.014 tell you about how much information each bounding method uses?
4. Why is the symmetry of the uniform distribution around its mean important for the Chebyshev calculation here?