All My Bloody Notes Bloody Maths Statistics Continuous Distribution

Continuous Distribution

What values will a random variable take


  1. Consider an random variable XX, which can take on a value in [0,w][0, w]. [0,w][0, w] is often referred to as the support of the distribution.
  2. Cumulative distribution: F(x)=Prob[Xx]F(x) = Prob[X \leq x]. We assume f(x)0f(x) \geq 0, for x[0,w]\forall x \in [0, w]
  3. Expectation: E[X]=0wxf(x)dxE[X] = \int_{0}^{w} x f(x) dx. Alternatively, as f(x)=dF(x)dxf(x) = \frac{dF(x)}{dx}, we can express the expectation as E[X]=0wxdF(x)E[X] = \int_{0}^{w} x dF(x)
  4. A degenerate random variable: a random variable that has a single possible value. Naturally, the variance of a degenerate variable is 0.

Conditional expectation

Contrast the conditional expectation of XX given XxX \leq x with the normal version:

Normal: E[X]=0wxf(x)dx=0wxdF(x)E[X] = \int_{0}^{w} x f(x) dx= \int_{0}^{w} x dF(x)

Conditional: E[Xx]=1F(x)0xtf(t)dtE[X \leq x] = \frac{1}{F(x)}\int_{0}^{x} t f(t) dt

Which can be viewed as a 2-step approach:

  1. Change the top limit
  2. Adjust the whole thing by multiplying 1F(x)\frac{1}{F(x)}

There is a graphical representation too that can be found in Krishna (2003)'s Book: "Auction Theory" in Appendix A:

F(x)E[XXx]=0xtf(t)dt=xF(x)0xF(t)dtF(x) E[X | X \leq x] = \int_{0}^{x}tf(t)dt = xF(x) - \int_{0}^{x}F(t)dt

The second equality is derived using integration by parts.

Intuition behind

Consider again the problem we encountered just now, E[XXx]E[X | X \leq x] can be written as:

E[Xx]=tf(ttx)dtE[X \leq x] = \int_{-\infty}^{\infty} t f(t | t \leq x) dt

Notice that we can break the conditional distribution into: E[Xx]=tf(t)1(t<x)F(x)dt=1F(x)xtf(t)dtE[X \leq x] = \int_{-\infty}^{\infty} t \frac{f(t)_{1(t < x)}}{F(x)} dt = \frac{1}{F(x)} \int_{-\infty}^{x} t f(t) dt

The reason is that the indicator function is always 00 for t>xt > x, and always 11 for txt \leq x. Therefore, it is the same as truncating the top limit to xx precisely because any tt above xx will have f(t)1(t<x)=0f(t)_{1(t < x)} = 0.

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