Conditional expectation of exponential distribution.

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SUMMARY

The discussion focuses on calculating the conditional expectation E[Y-X|Y>X] for two exponential distributions, X and Y, with means of 6 and 3, respectively. The correct formula derived is E[t_2-t_1|t_2>t_1] = E[(t_2-t_1)*I_{t_2>t_1}] = ∫₀^∞ ∫_{t_1}^∞ (t_2-t_1)λ₁λ₂e^{-λ₁t₁}e^{-λ₂t₂} dt₂ dt₁, leading to the result of λ₁/(λ₂(λ₁+λ₂)). The initial confusion arose from an error in integration by parts.

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manish13
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I have been stuck at this calculation. There are two exponential distributions X and Y with mean 6 and 3 respectively. We need to find
E[y-x|y>x]
I keep getting it negative, which is clearly wrong. Anybody wants to try it?
 
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Show your calculation. It should be easy to find the flaw.
 
I got it finally. But here is a brief note about it

[tex]E[t_2-t_1|t_2>t_1] = E[(t_2-t_1)*I_{\{t_2>t_1\}}] = \int_0^{\infty} \int_{t_1}^{\infty}(t_2-t_1)\lambda_1 \lambda_2 e^{-\lambda_1 t_1} e^{-\lambda_2 t_2} dt_2 dt_1[/tex]

This nicely gives the answer as [tex]\frac{\lambda_1}{\lambda_2(\lambda_1+\lambda_2)}[/tex]

I was doing a mistake in byparts.
 

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