Limits for a truncated random variable

[kobe87]

Suppose that X is a random variable distributed in the interval [a;b] with pdf f(x) and cdf F(x). Clearly, F(b)=1. I only observe X for values that are bigger than y.

I know that $E(X|X>y)=\frac{\int_y^b xf(x)dx}{1-F(y)}$.

Moreover, $\frac{∂E(X|X>y)}{∂y}=\frac{f(y)}{1-F(y)}[E(X|X>y)-y]$

I would like to evaluate this derivative as y→b.I was trying with Hopital but I could not go anywhere. Looking at Wikipedia(http://en.wikipedia.org/wiki/Truncated_distribution#Expectation_of_truncated_random_variable) it appears that the solution to my question is 1/2 but I might be completely wrong.

I am new to this forum I hope that I opened this thread in the right section. Thanks to anyone who is willing to help me.

 

[Ray Vickson]

Suppose that X is a random variable distributed in the interval [a;b] with pdf f(x) and cdf F(x). Clearly, F(b)=1. I only observe X for values that are bigger than y.

I know that $E(X|X>y)=\frac{\int_y^b xf(x)dx}{1-F(y)}$.

Moreover, $\frac{∂E(X|X>y)}{∂y}=\frac{f(y)}{1-F(y)}[E(X|X>y)-y]$

I would like to evaluate this derivative as y→b.I was trying with Hopital but I could not go anywhere. Looking at Wikipedia(http://en.wikipedia.org/wiki/Truncated_distribution#Expectation_of_truncated_random_variable) it appears that the solution to my question is 1/2 but I might be completely wrong.

I am new to this forum I hope that I opened this thread in the right section. Thanks to anyone who is willing to help me.

This is the right section.

The answer to your question depends on the form of f. Because I find it easier to deal with, I look instead at the equivalent problem $M(y) = E(X|X<y) \text{ and } \partial{M(y)}/\partial{y},$ for X on [0,a], a ≤ ∞; we want limits as y → 0+. Do you agree that is essentially the same problem?

Anyway, for two simple cases with f(x) discontinuous at x = 0 (viz., X ~ Uniform(0,a) and X ~ expl(rate=a)) I do get 1/2 as the limit of M'(y). However, for two cases where f(x) is continuous at x = 0, I get different answers. For f(x) = 2x/a^2, 0 < x < a, we have M(y) = (2/3)y, while for f(x) = 3x^2/a^3, 0 < x < a, we have M(y) = (3/4)y.

RGV

 

[kobe87]

Thanks for the reply. I agree that looking at the limit of M(y) as y→0⁺ is the same problem as well as the fact that the limit is 1/2 for the cases you mentioned. For the problem that I have in mind it is fine that the result of the limit depends on the shape of the distribution. However, I still need to have a solution for the general problem that tells me how this limit is affected by the shape of the distribution. Did you try to solve it under a general f(x)?
Thanks.

 

[kobe87]

I did the proof. It is 1/2 For any distribution

 

[Ray Vickson]

I did the proof. It is 1/2 For any distribution

That is not correct. For f1(x) = 2x/a^2, 0 < x < a, we have M1(y) = (2/3)y, while for f2(x) = 3x^2/a^3, 0 < x < a, we have M2(y) = (3/4)y. So, for f1 we have dM1/dy = 2/3, while for f2 we have dM2/dy = 3/4 near y = 0. These results for M1(y) and M2(y) are calculated using the formula
$$M(y) \equiv E(X|X<y) = \frac{\int_0^y x f(x) \, dx}{F(y)}, \; F(y) = \int_0^y f(x) \, dx.$$
Just go ahead and use these formulas for f = f1 and for f = f2.


In fact, you will get the limit 1/2 for any X that has f(x) discontinuous at x = 0 (that is, if f(x) = 0 for x < 0 but f(x) > 0 for x → 0+). However, it is most definitely NOT true for X that have f(x) continuous at x = 0, as is shown by the two examples above.

RGV

 

[kobe87]

Yeah it's true. In my previous post I was assuming discontinuity of the pdf in my proof. Thanks for the help.