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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.

difficulty advanced
Guest Jun 21, 2015

Best Answer 

 #1
avatar+248 
+5

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 =)

Brodudedoodebrodude  Jun 21, 2015
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1+0 Answers

 #1
avatar+248 
+5
Best Answer

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 =)

Brodudedoodebrodude  Jun 21, 2015

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