How to Calculate the Expected Value of X Given Y=5?

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SUMMARY

The discussion focuses on calculating the expected value of the random variable X given that Y equals 5, using the joint probability density function (pdf) f(x,y) = 1/50 for 0 < x < 10, 0 < y < 10, and 0 < x + y < 10. The correct approach involves determining the marginal pdf f(y) and the conditional expectation E[X|Y]. Participants clarify that the limits of integration for calculating E[X|Y=5] should be from 0 to 5, not 0 to 10. The final expected value calculation yields E[X|Y=5] = 10, but this result is questioned and requires careful verification of each step in the integration process.

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Homework Statement


A machine consists of 2 components whose lifetimes are X and Y and have joint pdf,
f(x,y)=1/50 w/ 0&lt;x&lt;10, 0&lt;y&lt;10,0&lt;x+y&lt;10
Calculate the expected value of X given Y=5.

Homework Equations


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


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.
 
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mynameisfunk said:

Homework Statement


A machine consists of 2 components whose lifetimes are X and Y and have joint pdf,
f(x,y)=1/50 w/ 0&lt;x&lt;10, 0&lt;y&lt;10,0&lt;x+y&lt;10
Calculate the expected value of X given Y=5.

Homework Equations


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


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.

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).
 
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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 can't get this latex to work but I am getting 10 for my answer. Does that seem right?
 
mynameisfunk said:
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 can't get this latex to work but I am getting 10 for my answer. Does that seem right?

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.
 

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