Let r be a continuous random variable governed by a Gaussian PDF with parameters m and v .Show that E(r)=m and V(r)=v.

Question

In your derivation you can use the results that 1/(√2πv)∫^+∞_−∞ e^{−(x−m)^2/2v} dx=1

 

-> I have the solution but I am unsure about the following:

 

 the solution says E(r) = 1/(√2πv)∫^+∞−∞ xe^{−(x−m)^2/2v} dx, but I don't know why;

 

it then goes on to say: = v/(√2πv)∫^+∞_−∞ (x-m + m)/v . e^{−(x−m)^2/2v} dx , and I am also unsure as to how they get this from the line above.

 

It goes on to splitting it into 2 integrals and eventually solving to get m which I can just about follow.

Answer 1

You just have to remember the definition of E[X] and Var[x]:

E[X]=∫xp(x)dx, where p(x) is the PDF

Var[X]=E[X²]-(E[X])²=∫x²p(x)dx-(∫xp(x)dx)²

Now your PDF is normal with m and v. So if you substitute normal PDF into equations from the definition you'll get formulas in your question.

E(r)=1/(√2πv)∫_-∞^+∞xe^{−(x−m)^2/2v}dx. You have to evaluate this integral (you should get m) and the one for variance as well. You can do it by integration by parts. in that case, you will use the fact that 1/(√2πv)∫p(x)dx=1.