Why am I getting the wrong result for the expected value of an F-distribution?

However, this is not generally true. In fact, in this case, E(1/W2) = v2/(v2-2).In summary, the expected value of an F-distributed random variable is not simply 1, as assumed in the conversation. The hiccup in the logic is assuming that E(1/W2) = 1/E(W2), when in fact it is equal to v2/(v2-2).
  • #1
thisguy12
3
0
Let W_1 and W_2 be independent Chi-Squared distributed random variables with v_1 and v_2 degrees of freedom, respectively. Then F = (W_1/v_1)/(W_2/v_2) = (v_2/v_1)(W_1/W_2) is said to have an F-distribution with v_1 numerator degrees of freedom and v_2 denominator degrees of freedom.

I want to find E(F).

Here is where I get confused:
E(F) = E((v_2/v_1)(W_1/W_2))
= (v_2/v_1) E(W_1/W_2) since v_1, v_2 are constants
= (v_2/v_1) E(W_1)/E(W_2) because W_1 and W_2 are independent random variables
= (v_2/v_1) (v_1/v_2) because if X is chi-squared r.v. with v degrees of freedom, E(X) = v
= 1


However, the correct result for the expected value of an F-distributed r.v. is v_2/(v_2 - 2). Where is the hiccup in my logic?
 
Physics news on Phys.org
  • #2
You are assuming E( 1/W2) = 1/ E(W2).

If W is a random variable with density f(w), this would say
[itex] \int \frac{1}{w} f(w) dw [/itex] = [itex] \frac{1}{\int f(w) dw} [/itex]
 

1. Why is the expected value of an F-distribution not equal to its degrees of freedom?

The expected value of an F-distribution is defined as the ratio of the degrees of freedom in the numerator and denominator. However, this does not mean that the expected value will always be equal to the degrees of freedom. The expected value represents the long-term average of many F-distribution samples, while the degrees of freedom represent the parameters used to calculate the F-value for a specific sample. Thus, they are not equivalent.

2. Can the expected value of an F-distribution be negative?

No, the expected value of an F-distribution can never be negative. This distribution is always positive and skewed to the right. A negative expected value would imply that the F-distribution has a negative mean, which is not possible.

3. How does the sample size affect the expected value of an F-distribution?

The sample size does not directly affect the expected value of an F-distribution. However, as the sample size increases, the F-distribution will converge to the theoretical expected value. This is because a larger sample size leads to a more accurate estimation of the population parameters, leading to a more stable expected value.

4. Can the expected value of an F-distribution be greater than 1?

Yes, the expected value of an F-distribution can be greater than 1. This is because the F-distribution is a continuous probability distribution, and its expected value represents the long-term average of many samples. Therefore, the expected value can take on any value within its range, including values greater than 1.

5. Why is the expected value of an F-distribution important?

The expected value of an F-distribution is important because it represents the central tendency of the distribution. It is used to calculate other statistical measures such as the variance and standard deviation, which are essential in hypothesis testing and making statistical inferences. Additionally, the expected value can also help interpret the results of an F-test and determine the significance of the relationship between variables in an ANOVA analysis.

Similar threads

Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
894
  • Set Theory, Logic, Probability, Statistics
Replies
14
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
Replies
0
Views
335
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
Replies
2
Views
638
Back
Top