Hi Not really grasped the concept of randomized strategies. How does choosing a randomized strategy by any player ensures maximization or optimization of pay offs? When we are choosing a random strategy, we are leaving the course of action to randomness or chance. Right? So how does it ensure reduction of expected maximum loss to a minimum? Under the topic 'Statistical Games' on page 19 of chapter 1, an example is given in which risk function for each of the decision function is to be calculated. And the example is- A statistician is observing values from a Bin(2, p) distribution. He knows that p is equal either to ¼ or to ½, and he is trying to choose between these two values. He observes a single value x from the distribution. He proposes to use one of the following four decision functions: d1(x) : Set p=¼ when x=0 Set p=½ when x= 1 or 2 d2 (x) : Set ¼ when 0 or 1 Set ½ when x= 2 d3(x) : Set p =¼ when x = 0, 1 or 2 d4 (x) : Set p =½ when x = 0, 1 or 2 If he incorrectly concludes that p = ¼ , he suffers a loss of 1. If he incorrectly concludes that p = ½, he suffers a loss of 2. Find the risk function for each decision function, and find the decision function that minimises the maximum expected loss. NOT UNDERSTOOD THE QUESTION COMPLETELY- What are these values of x=0, 1 or 2??? In calculation, the risk function has a component.... Consider first the decision function d1(x) . If p actually is ¼, there is a loss of 0 if x = 0 and a loss of 2 if x = 1 or 2 . So: R(d1,p=1/4 ) = 0* P(x=0 | p=1/4) + 2 * P(X=1 or 2 | p= 1/4 ) = 2*[2* 1/4 * 3/4 + (1/4) ^2] = 7/8 How this second calculation has been arrived at???? 2 * P(X=1 or 2 | p= 1/4 ) = 2*[2* 1/4 * 3/4 + (1/4) ^2] = 7/8 REQUEST A REVERT ASAP... BTW, ANYONE FROM MUMBAI APPEARING FOR CT6 in MAY DIET 2014? OR SOMEONE WHO CLEARED THE EXAM AND READY TO HELP? Regards
how risk function has been arrived at? Ok. I have understood the calculation of p(x= 1 or 2 | p=1/4). But not really not understood how risk function has been arrived at?
\(R(d_1, p = \frac{1}{4})\) means Expected Loss if the parameter was actually \(\frac{1}{4}\) and statistician went with Decision function 1 Suppose parameter(p) is actually \(\frac{1}{4}\) Now under \(d_1(x)\) statistician's decision is to say: \(p = \frac{1}{4}\) if he observes a 0 and \(p = \frac{1}{2}\) if he observes either 1 or 2 That means he will suffer a loss of 2 if either 1 or 2 is generated from Bin(2,¼) because he'll wrongly conclude that \((p = \frac{1}{2})\) when in fact it was equal to \(\frac{1}{4}\) and P[1 or 2 is generated from Bin(2, ¼)] = \(1 - (\frac{3}{4})^2 = \frac{7}{16}\) Hence he'll suffer a loss of 2 with probability 7/16 So, \(R(d_1, p = \frac{1}{4}) = 2 \times \frac{7}{16}\) = \(\frac{7}{8}\)
