kwy said:Homework Statement
EX=EY=5, VarX=1, VarY=sigma^2 >1
Z=aX+(1-a)Y, 0<=a<=1
find a that minimizes VarZ, and another a that maximize VarZ
Homework Equations
Yes, to the last question. In fact, Var(a*X + b*Y) = a^2 * Var(X) + b^2*Var(Y) + 2*a*b*Cov(X,Y) for constants a and b.kwy said:Thanks for your patience, the course went so fast, I am still trying to revise all the notes between full time work.
so Var(W)= k^2Var(P)
does Var(W+Q)= VarW + VarQ + 2xCov(W,Q)??
kwy said:Hi Ray, since the RVs X and Y are independent, I don't have to worry about the Cov(XY) right? Hence, the equation is now
VarZ = a^2VarX + (1-a)^2VarY (VarX = 1, VarY = sigma^2)
= a^2 + (1-a)^2*sigma^2 and where do the mean values come in?
How can I differentiate this?
Thanks in advance.
kwy said:Hi Ray, since the RVs X and Y are independent, I don't have to worry about the Cov(XY) right? Hence, the equation is now
VarZ = a^2VarX + (1-a)^2VarY (VarX = 1, VarY = sigma^2)
= a^2 + (1-a)^2*sigma^2 and where do the mean values come in?
How can I differentiate this?
Thanks in advance.
Your function f(a) is a^2 + v*(1-a)^2, where v = sigma^2. The graph y = f(a) is an upward-opening parabola on the whole real a-line, so on 0 <= a <= 1, only three possibilities can occur: (i) f(a) is decreasing (i.e., non-increasing) as a function of a on the interval [0,1]; (ii) f(a) is increasing on [0,1]; or (iii) f(a) is decreasing on the left part of [0,1] and increasing on the right part. (Draw sketches to pin down your understanding.) Now you need to determine whether you are in case (i), (ii) or (iii), and then determine how to find the max and min of f(a) on the a-interval [0,1].kwy said:Hi Guys, I think the means may be of use somewhere. There is no 2nd part to the question, the aim is to determine values of a that will i) minimise VarZ and ii) maximise VarZ. Following differentiation, a = sigma^2/(1+sigma^2). From this value and 2 random variables, how can I check the "endpoint extema"? Do I simply test with F(0,0) and F(1,1). Many thanks.
The purpose of using probability theory in mini max functions is to quantify the likelihood of different outcomes in a game or decision-making scenario. It allows us to assign numerical values to uncertain events and make informed decisions based on the probabilities of different outcomes.
Probability theory helps in minimizing risk in mini max functions by providing a framework for calculating the expected value of different outcomes. By calculating the expected value, we can choose the option that minimizes potential losses and maximizes potential gains.
Yes, probability theory can be applied to all types of mini max functions as long as there is an element of uncertainty involved. It can be applied to various fields such as economics, game theory, and decision analysis.
The concept of expected value is closely related to mini max functions as it involves calculating the average outcome of a decision or strategy over a large number of trials. In mini max functions, the goal is to minimize the maximum potential loss, and the expected value helps in making the best decision to achieve this goal.
One limitation of using probability theory in mini max functions is that it assumes that all outcomes are equally likely to occur. In real-world scenarios, this may not always be the case, and the calculated probabilities may not accurately reflect the actual outcomes. Additionally, probability theory does not take into account personal biases or preferences, which may also impact decision-making in mini max functions.