Simple doubt on Expectations (Probability)

[knowpraveen]

How is this possible? Could anyone solve this for me? I got x/2 as the answer. Moreover, isn't it possible to eliminate other options based on the fact that the expectations wouldn't be greater than 1 since x lies in the interval 0 and 1- which conveniently makes the last three options invalid. Where have I gone wrong? Thanks.

[varnika1880]

Even m getting it as X/2 only ... .can u plz tell which formula has been used in the answer given ...

[knowpraveen]

By me?

E(X|X<=x)

So, lets say F(y)= P(X<=y|X<=x) = y/x.
Therefore, the density would be f(y)=1/x

The expectation would be integral(y(1/x)) from 0 to x. Hence, the answer comes to be at x/2.

The answer is consistent with intuition, because, for X less than or equal to x, expectation should be the mean of x and 0 i.e x/2. Or, the expectation shouldn't be greater than 1 since x lies between 0 and 1. For all other options, the expectations might exceed 1 for some values of x belongs to (0,1)

What do you say about this?

[Amit Goyal]

It is x/2 because X|(X <= x) ~ U(0,x)

[Qwerty]

knowpraveen from which year question paper is this??

[knowpraveen]

Thank you, sir. Qwerty, this is from DSE Super 20 mock paper.