# Expected Value (joint pdf)

1. Nov 30, 2013

### mynameisfunk

1. The problem statement, all variables and given/known data
A machine consists of 2 components whose lifetimes are X and Y and have joint pdf,
$$f(x,y)=1/50$$ w/ $$0<x<10$$, $$0<y<10$$,$$0<x+y<10$$
Calculate the expected value of $$X$$ given $$Y=5$$.

2. Relevant equations
$$E[X|Y]= \int_{-inf}^{inf} x f(x,y)/f(y) dx$$

3. The attempt at a solution
$$f(y) = \int_{0}^{10-y} 1/50 dx = (1-y)/5$$

$$E[x|y] = \int_{0}^{10-y} x/(10-10y) dx$$

This is where I am confused. How do i set the limits on the integral where I actually compute the expectation? Is x=0 to x=10-y right? I was thinking I want to just integrate as normal and plug in y=5 after the integration.

2. Nov 30, 2013

### Ray Vickson

Careful:
$$f(y) = \int_{0}^{10-y} 1/50 dx = \frac{10-y}{50} \neq (1-y)/5 \leftarrow \text{ what you wrote}$$

So, you should have a different $f(x|y)$ from what you wrote, which means that $E(X|Y=y)$ will be different from what you wrote. And, finally, YES, you should put y = 5 after the integration (or even before the integration, since y is kept constant throughout).

3. Nov 30, 2013

### mynameisfunk

thank you!

OK, thanks for that. OK, just to be clear, can you tell me if this looks right?

[ itex ]E[X|Y=5] = \int_{0}^{10} x E[X|Y=5] dx \int_{0}^{1} x * 1/5 dx = \left[ \frac{10}{2}x^2 \right]_{0}^{10} = 10[ /itex ]

Not sure why I cant get this latex to work but I am getting 10 for my answer. Does that seem right?

4. Nov 30, 2013

### Ray Vickson

Edited version of the above:
$$E[X|Y=5] = \int_{0}^{10} x E[X|Y=5] dx \int_{0}^{1} x * 1/5 dx = \left[ \frac{10}{2}x^2 \right]_{0}^{10} = 10$$
(obtained by replacing "[ itex ]" by "[tex ]" (remove the space between 'x' and ']' and replacing "[ /itex ]" by "[/tex ]"', again removing the final space. If you want in-line formulas (such as produced by "[itex ]"---no spaces) it is easier to use "# #" (no space between the two #s) at each end of the formula.

Anyway, your formula is incomprehensible to me, and I cannot figure out why you would ever assume it is correct. It is essentially saying (for $A = E[X|Y=5]$) that
$$A = \int_0^{10} A x \, dx \cdot \int_0^1 \frac{x}{5} dx.$$
and that says
$$A = A \frac{100}{2} \cdot \frac{1}{2} \frac{1}{5} = 50 A,$$
which could only be true for A = 0.

Go back to square one, and go carefully. Just take your time, and check every step.