finding expected value

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• Nov 19th 2009, 05:44 AM
noob mathematician
finding expected value
Let X_1,....,X_n be iid rv with density f and cumulative distribution function F. Let:

$\displaystyle I_{X1}(a)=1, (\text{if } X_1\leq a) \text{ and}=0, \text{ (otherwise)}$

I want to find the expected value of $\displaystyle I_{X1}(2)$

I can't seems to find the answer since the distribution is unknown to me
• Nov 19th 2009, 08:36 PM
matheagle
Quote:

Originally Posted by noob mathematician
Let X_1,....,X_n be iid rv with density f and cumulative distribution function F. Let:

$\displaystyle I_{X1}(a)=1, (\text{if } X_1\leq a) \text{ and}=0, \text{ (otherwise)}$

I want to find the expected value of $\displaystyle I_{X1}(2)$

I can't seems to find the answer since the distribution is unknown to me

The expected value of an indicator function is just the probability of that event.

Hence $\displaystyle E(I_{X1}(2))=P(X1\le 2)=F_{X1}(2)$
• Nov 19th 2009, 10:13 PM
noob mathematician
Suppose now $\displaystyle p=P(X1\leq 2).$ How to find the mle (maximum likelihood estimate) of p, based on the whole sample.
• Nov 19th 2009, 10:20 PM
matheagle
I would think that these are i.i.d. Bernoulli's.

Let $\displaystyle Y_i=I(X_i\le 2)$ for i=1,2,...,n.

Here $\displaystyle p=P(X1\le 2)$ isn't really the point

The MLE of Bernoullis is $\displaystyle \hat p$ the sample mean of the Y's.

$\displaystyle \hat p={\sum_{i=1}^nI(X_i\le 2)\over n}$

Since the Likelihood function is

$\displaystyle L= p^{Y_1}(1-p)^{1-Y_1}p^{Y_2}(1-p)^{1-Y_2}\cdots p^{Y_n}(1-p)^{1-Y_n}$

$\displaystyle = p^{\sum_{i=1}^n Y_i}(1-p)^{n-\sum_{i=1}^n Y_i}$

Now take the log and differentiate TWICE and obtain the sample mean as the MLE.
• Nov 19th 2009, 11:03 PM
noob mathematician
Given the Y_i defined by u:

Let say now I want to know the MLE of p(1-p), is it right for me to conclude that it is $\displaystyle \overline{Y}(1-\overline{Y})$?
• Nov 19th 2009, 11:12 PM
matheagle
I believe that is correct under the invariance principle of MLEs.