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Cumulative Distribution of a Linear Transformation of a R.V

Hello guys, I just needed help on this question below.

Screen Shot 2018-12-26 at 18.47.46.png

The question specifies that if the device fails before T = 3, i.e. it has a lifetime between 0 <= t < 3, then V = 5.
Otherwise, it has value 2T. So, am I correct to assume V = 2T if t > 3 i.e. 3 < t < infinity (or upper bound of t which is unknown)?

Also, would it be incorrect to put the upper bound of the integral as infinity?
(edited 5 years ago)
Reply 2
Original post by Chittesh14
Hello guys, I just needed help on this question below.

Screen Shot 2018-12-26 at 18.47.46.png

The question specifies that if the device fails before T = 3, i.e. it has a lifetime between 0 <= t < 3, then V = 5.
Otherwise, it has value 2T. So, am I correct to assume V = 2T if t > 3 i.e. 3 < t < infinity (or upper bound of t which is unknown)?

Also, would it be incorrect to put the upper bound of the integral as infinity?


The upper "limit" of the exponential pdf is infinity, so it sounds like you're interpreting it correctly.
The CDF of T is fairly straightforward, have you calculated it? Then think about how it is transformed to V
Reply 3
Original post by mqb2766
The upper "limit" of the exponential pdf is infinity, so it sounds like you're interpreting it correctly.
The CDF of T is fairly straightforward, have you calculated it? Then think about how it is transformed to V


Yes, I've calculated the cdf of T. I didnt know how to transform it to V, but I think I know. What would the new upper limit be if I transformed it to V?
Reply 4
Original post by Chittesh14
Yes, I've calculated the cdf of T. I didnt know how to transform it to V, but I think I know. What would the new upper limit be if I transformed it to V?

Should be infinity again as V=2T? Generally, you'd just say T>0 as the question implies.

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