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.
+250 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)]=∫∞−∞[ag(X)+bh(X)]fx(x)dx\indent=∫∞−∞ag(X)fx(x)dx+∫∞−∞bh(X)fx(x)dx\indent=a∫∞−∞g(X)fx(x)dx+b∫∞−∞h(X)fx(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=∫∞−∞a2X2fx(x)dx−(∫∞−∞aXfx(x)dx)2\indent=a2∫∞−∞X2fx(x)dx−(a∫∞−∞Xfx(x)dx)2\indent=a2E[X2]−a2(∫∞−∞Xfx(x)dx)2\indent=a2(E[X2]−(E[X])2)\indent=a2Var(X)
I'm not too sure with part c so hopefully someone else can help with that =)
+250 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)]=∫∞−∞[ag(X)+bh(X)]fx(x)dx\indent=∫∞−∞ag(X)fx(x)dx+∫∞−∞bh(X)fx(x)dx\indent=a∫∞−∞g(X)fx(x)dx+b∫∞−∞h(X)fx(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=∫∞−∞a2X2fx(x)dx−(∫∞−∞aXfx(x)dx)2\indent=a2∫∞−∞X2fx(x)dx−(a∫∞−∞Xfx(x)dx)2\indent=a2E[X2]−a2(∫∞−∞Xfx(x)dx)2\indent=a2(E[X2]−(E[X])2)\indent=a2Var(X)
I'm not too sure with part c so hopefully someone else can help with that =)
