Calculate the expected value for filling up a cup

[Addez123]

Homework Statement

Into a cup with volume 'a' you pour volume X.
X has density function
$$fx(x) = (x+1)^{-2}, x \geq 0$$
Y = volume in the cup after you've poured in X.
Calculate E(Y)
Note: if x > a then the cup flows over.

Relevant Equations

$$E(Y) = \int g(x) * fx(x) dx, Y = g(x)$$

My guess is that g(x) = x?
The limits of integration should be 0 to a, since after a the cup flows over.

If I put these in, I get the solution (I've doubled checked with wolfram alpha that it's correct):
$$E(Y) = ln(a+1) + 1/(a+1) - 1$$

The textbook solution is just ln(a+1).
I'm super new to the whole concept of probability so I'm just trying to make qualified guesses at what all the variables mean, obviously I did something wrong but what?

 

[member 587159]

So, we start with a cup with volume $a$ and then we add volume $X$ to the cup. We end up with a volume of $a+X$, so we get $Y = a + X$. Thus $Y = g(X)$ where $g(x) = a+x$.

Now, $$\mathbb{E}(Y) = \int g(x) f_X(x) dx = \int (a+x) (x+1)^{-2}dx$$ should give the right answer.

 

[Addez123]

So, we start with a cup with volume $a$ and then we add volume $X$ to the cup. We end up with a volume of $a+X$, so we get $Y = a + X$. Thus $Y = g(X)$ where $g(x) = a+x$.

Now, $$\mathbb{E}(Y) = \int g(x) f_X(x) dx = \int (a+x) (x+1)^{-2}dx$$ should give the right answer.

I think the cup is empty, the actual volume of the cup is a. Cause it says: if x > a then the cup runs over)

 

[member 587159]

I think the cup is empty, the actual volume of the cup is a. Cause it says: if x > a then the cup runs over)

Ok, maybe post the entire problem statement next time? I misinterpreted: Here is what you do:

You start pouring a volume $X$ into an empty cup with total volume $a$. Thus the total volume in the cup is $$Y(\omega):= \begin{cases}X(\omega) \quad X(\omega) \leq a \\ a \quad X(\omega) \geq a\end{cases}$$
Thus $Y= g(X)$ where
$$g(x) = \begin{cases}x \quad x \leq a \\ a \quad x \geq a\end{cases}$$
Hence,
$$\mathbb{E}(Y) = \int_0^\infty g(x) f_X(x) dx = \int_0^a x (x+1)^{-2}dx + a\int_a^\infty (x+1)^{-2}dx$$

Does this give you the correct result? (Wolfram alpha gives $\ln(a+1)$ as a result for this calculation, which corresponds to your textbook answer).

 

[Addez123]

Haha I'm sorry, I read it and thought it was a very unnessecary detail to add, obviously I was wrong :p
Yes, if I integrate from a to inf, with the condition that g(x)=a, it balance out and I get correct answer. Thanks!