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# Stats Proofs

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I know these results however I've never really learnt how to prove these questions:

Assume that X is a continuous random variable that takes on values in the set of real numbers. Show that the following equalities hold:

a) For any functions g and h and constants a and b: E[ag(X)+bh(X)]=aE[g(X)]+bE[h(X)]

b) Var(aX)=a^2Var(X), where a is again a constant

c) E[E(X|Y=y)] = E(X), with Y being another random variable. The "outer" expectation is taken over values of Y, and the "inner" one is over the values of X.

Jun 21, 2015

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The 'tricks' for the first one is that you can take a constant outside of an integral and also that you can split up a single integral of many terms into many integrals of single terms:

$$E[ag(X)+bh(X)]= \int_{-\infty}^{\infty} [ag(X)+bh(X)]f_x(x)dx \\\indent= \int_{-\infty}^{\infty}ag(X)f_x(x)dx+\int_{-\infty}^{\infty}bh(X)f_x(x)dx \\\indent =a\int_{-\infty}^{\infty}g(X)f_x(x)dx+b\int_{-\infty}^{\infty}h(X)f_x(x)dx \\\indent = aE[g(X)]+bE[h(x)]$$

Part b is a very similar process to part a:

$$Var(aX) = E[(aX)^2]-(E[aX])^2 \\\indent =\int_{-\infty}^{\infty}a^2X^2f_x(x)dx-(\int_{-\infty}^{\infty}aXf_x(x)dx)^2\\\indent= a^2\int_{-\infty}^{\infty}X^2f_x(x)dx-(a\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2E[X^2]-a^2(\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2(E[X^2]-(E[X])^2)\\\indent=a^2Var(X)$$

I'm not too sure with part c so hopefully someone else can help with that =)

Jun 21, 2015

#1
+5
$$E[ag(X)+bh(X)]= \int_{-\infty}^{\infty} [ag(X)+bh(X)]f_x(x)dx \\\indent= \int_{-\infty}^{\infty}ag(X)f_x(x)dx+\int_{-\infty}^{\infty}bh(X)f_x(x)dx \\\indent =a\int_{-\infty}^{\infty}g(X)f_x(x)dx+b\int_{-\infty}^{\infty}h(X)f_x(x)dx \\\indent = aE[g(X)]+bE[h(x)]$$
$$Var(aX) = E[(aX)^2]-(E[aX])^2 \\\indent =\int_{-\infty}^{\infty}a^2X^2f_x(x)dx-(\int_{-\infty}^{\infty}aXf_x(x)dx)^2\\\indent= a^2\int_{-\infty}^{\infty}X^2f_x(x)dx-(a\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2E[X^2]-a^2(\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2(E[X^2]-(E[X])^2)\\\indent=a^2Var(X)$$