+ Reply to Thread
Results 1 to 3 of 3

Thread: Second Order Stochastic Dominance

  1. #1
    Points: 1,510, Level: 22
    Level completed: 10%, Points required for next Level: 90

    Posts
    2
    Thanks
    1
    Thanked 0 Times in 0 Posts

    Second Order Stochastic Dominance




    how can I test for second order stochastic dominance? is easy to understand the idea but when I have to normal distributions with different means and different standard deviations I don't know how to calculate if the integrals of the CDF intersect

  2. #2
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Second Order Stochastic Dominance

    Suppose you want to check whether X_1 has a second order stochastic dominance over X_2. So you would like to consider the following function:

    D(x) = \int_{-\infty}^{x} [F_{X_2}(t) - F_{X_1}(t)] dt

    and check whether D(x) \geq 0 ~~ \forall x \in \mathbb{R}

    If their CDF is nice enough, D will be differentiable and you can try to locate the local minimum points of D, and check whether they are all non-negative. By differentiation, it is easy to see that all critical point(s) x need to satisfy

    F_{X_1}(x) = F_{X_2}(x)

    and for those critical points will be either local maximum or local minimum.

    In particular, if you assume that X_1 \sim (\mu_1, \sigma^2_1) and X_2 \sim (\mu_2, \sigma^2_2),

    F_{X_1}(x) = F_{X_2}(x) \iff \Pr\{X_1 \leq x\} = \Pr\{X_2 \leq x\}

    \iff \Pr\left\{Z \leq \frac {x - \mu_1} {\sigma_1} \right\}= \Pr\left\{Z \leq \frac {x - \mu_2} {\sigma_2} \right\}

    \iff \frac {x - \mu_1} {\sigma_1} = \frac {x - \mu_2} {\sigma_2}

    \iff (\sigma_2 - \sigma_1)x = \sigma_2\mu_1 - \sigma_1\mu_2

    Now we see that if their variances are equal \sigma_1 = \sigma_2, their CDFs will not intersect each other (trivial case: CDFs are overlapping if their mean also equal). So you would like to check whether

    F_{X_2}(x) > F_{X_1}(x) \iff \frac {x - \mu_1} {\sigma} < \frac {x - \mu_2} {\sigma} \iff \mu_1 > \mu_2

    as expected.

    When the variances are not equal \sigma_1 \neq \sigma_2, the above equation will give one and only one root. That means the function D have one critical point only:

    x^* = \frac {\sigma_2\mu_1 - \sigma_1\mu_2} {\sigma_2 - \sigma_1}

    So now you need to verify the following:

    1. F_{X_2}(x) - F_{X_1}(x) > 0 ~~ \forall x < x^*

    and this ensure D(x) > 0 ~~ \forall x < x^*. From the definition it ensure that D will be strictly increasing up to x^* and strictly decreasing until +\infty. And thus we would also need to verify

    2. \lim_{x\to+\infty}D(x) \geq 0

    For the first condition it can be similarly translated as

    (\sigma_2 - \sigma_1)x < \sigma_2\mu_1 - \sigma_1\mu_2 ~~ \forall x < x^*

    and this is equivalent to \sigma_2 > \sigma_1
    (as expected; a larger variance ensure a thicker tail)

    For the second condition, note that

    \lim_{x\to+\infty}D(x) = \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)] dt

    Therefore

    \lim_{x\to+\infty}D(x) \geq 0 \iff 
\int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)]dt \geq 0

    \iff \int_{-\infty}^{0} \{-F_{X_1}(t) - [-F_{X_2}(t)]\}dt 
+ \int_{0}^{+\infty} \{[1 - F_{X_1}(t)] - [1 - F_{X_2}(t)] \}dt \geq 0

    \iff \int_{-\infty}^{0} [-F_{X_1}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_1}(t)]dt
\geq \int_{-\infty}^{0} [-F_{X_2}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_2}(t)]dt

    \iff E[X_1] \geq E[X_2]

    As a result, the conditions will be mu_1 \geq \mu_2 and \sigma_1 \leq \sigma_2 and this should be enough for normal distribution.

    The issue becomes interesting if you have the two independent random sample and want to test the stochastic dominance. Again if you have the normality assumption, you may have the above condition as your null hypothesis.

  3. The Following 2 Users Say Thank You to BGM For This Useful Post:

    GretaGarbo (02-27-2013), santicanel (03-10-2013)

  4. #3
    Points: 1,510, Level: 22
    Level completed: 10%, Points required for next Level: 90

    Posts
    2
    Thanks
    1
    Thanked 0 Times in 0 Posts

    Thumbs up Re: Second Order Stochastic Dominance


    Sorry took me a while to answer. Thanks you very much. Really helped!

    Was reading Levy's book on stochastic dominance and remember my question

    Regards!

    Quote Originally Posted by BGM View Post
    Suppose you want to check whether X_1 has a second order stochastic dominance over X_2. So you would like to consider the following function:

    D(x) = \int_{-\infty}^{x} [F_{X_2}(t) - F_{X_1}(t)] dt

    and check whether D(x) \geq 0 ~~ \forall x \in \mathbb{R}

    If their CDF is nice enough, D will be differentiable and you can try to locate the local minimum points of D, and check whether they are all non-negative. By differentiation, it is easy to see that all critical point(s) x need to satisfy

    F_{X_1}(x) = F_{X_2}(x)

    and for those critical points will be either local maximum or local minimum.

    In particular, if you assume that X_1 \sim (\mu_1, \sigma^2_1) and X_2 \sim (\mu_2, \sigma^2_2),

    F_{X_1}(x) = F_{X_2}(x) \iff \Pr\{X_1 \leq x\} = \Pr\{X_2 \leq x\}

    \iff \Pr\left\{Z \leq \frac {x - \mu_1} {\sigma_1} \right\}= \Pr\left\{Z \leq \frac {x - \mu_2} {\sigma_2} \right\}

    \iff \frac {x - \mu_1} {\sigma_1} = \frac {x - \mu_2} {\sigma_2}

    \iff (\sigma_2 - \sigma_1)x = \sigma_2\mu_1 - \sigma_1\mu_2

    Now we see that if their variances are equal \sigma_1 = \sigma_2, their CDFs will not intersect each other (trivial case: CDFs are overlapping if their mean also equal). So you would like to check whether

    F_{X_2}(x) > F_{X_1}(x) \iff \frac {x - \mu_1} {\sigma} < \frac {x - \mu_2} {\sigma} \iff \mu_1 > \mu_2

    as expected.

    When the variances are not equal \sigma_1 \neq \sigma_2, the above equation will give one and only one root. That means the function D have one critical point only:

    x^* = \frac {\sigma_2\mu_1 - \sigma_1\mu_2} {\sigma_2 - \sigma_1}

    So now you need to verify the following:

    1. F_{X_2}(x) - F_{X_1}(x) > 0 ~~ \forall x < x^*

    and this ensure D(x) > 0 ~~ \forall x < x^*. From the definition it ensure that D will be strictly increasing up to x^* and strictly decreasing until +\infty. And thus we would also need to verify

    2. \lim_{x\to+\infty}D(x) \geq 0

    For the first condition it can be similarly translated as

    (\sigma_2 - \sigma_1)x < \sigma_2\mu_1 - \sigma_1\mu_2 ~~ \forall x < x^*

    and this is equivalent to \sigma_2 > \sigma_1
    (as expected; a larger variance ensure a thicker tail)

    For the second condition, note that

    \lim_{x\to+\infty}D(x) = \int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)] dt

    Therefore

    \lim_{x\to+\infty}D(x) \geq 0 \iff 
\int_{-\infty}^{+\infty} [F_{X_2}(t) - F_{X_1}(t)]dt \geq 0

    \iff \int_{-\infty}^{0} \{-F_{X_1}(t) - [-F_{X_2}(t)]\}dt 
+ \int_{0}^{+\infty} \{[1 - F_{X_1}(t)] - [1 - F_{X_2}(t)] \}dt \geq 0

    \iff \int_{-\infty}^{0} [-F_{X_1}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_1}(t)]dt
\geq \int_{-\infty}^{0} [-F_{X_2}(t)]dt + \int_{0}^{+\infty} [1 - F_{X_2}(t)]dt

    \iff E[X_1] \geq E[X_2]

    As a result, the conditions will be mu_1 \geq \mu_2 and \sigma_1 \leq \sigma_2 and this should be enough for normal distribution.

    The issue becomes interesting if you have the two independent random sample and want to test the stochastic dominance. Again if you have the normality assumption, you may have the above condition as your null hypothesis.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats