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

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Homework Help Overview

The discussion revolves around calculating the expected value of a random variable X given that another random variable Y equals 5. The context involves joint probability density functions (pdf) and the application of conditional expectations within the framework of probability theory.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the expected value E[X|Y=5] using the joint pdf provided. There are attempts to derive the marginal pdf f(y) and the conditional expectation E[X|Y]. Questions arise regarding the limits of integration and the correctness of the expressions used in the calculations.

Discussion Status

Some participants have provided corrections to earlier attempts, pointing out discrepancies in the calculations of f(y) and the resulting conditional expectation. There is an ongoing exploration of the correct setup for the integrals and the implications of the assumptions made in the problem.

Contextual Notes

Participants note potential confusion regarding the limits of integration and the proper formulation of the conditional expectation. There is also mention of issues with LaTeX formatting in the expressions shared, which may affect clarity in communication.

mynameisfunk
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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<x<10, 0<y<10,0<x+y<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<x<10, 0<y<10,0<x+y<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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