# Homework Help: Limits for a truncated random variable

1. Apr 17, 2012

### 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.

2. Apr 17, 2012

### Ray Vickson

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

3. Apr 18, 2012

### 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.

Last edited: Apr 18, 2012
4. Apr 18, 2012

### kobe87

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

5. Apr 18, 2012

### Ray Vickson

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

6. Apr 18, 2012

### kobe87

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