Probability Mass Function of { Z | Z < 1 }, Z given Z is less than 1

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SUMMARY

The discussion focuses on deriving the probability density function (PDF) of the random variable Z, defined as Z = X + Y, under the condition that Z is less than 1. The correct formulation is f(z) = P(Z=z|Z<1) = P(Z=z)/P(Z<1) for z < 1, with f(z) equating to 0 for z ≥ 1. The PDF can be expressed as f(z) = g(z)/∫_{-∞}^1 g(z) dz for z ≤ 1, where g(z) is the PDF of Z = X + Y, assuming X and Y are uniformly distributed over [0,1].

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Probability Mass Function of { Z | Z < 1 }, "Z given Z is less than 1"

Homework Statement



Given Z = X + Y.

Find the probability density function of Z|Z < 1.


Homework Equations



N.A.

The Attempt at a Solution



f(z) = P(Z=z|Z<1) = P(Z=z AND Z < 1) / P(Z < 1).

I thought the top could be simplified to P(Z=z) for z < 1. Correct?

So,

f(z) = 0, for z > or = 1
f(z) = P(Z=z)/P(Z<1), for z < 1.

Correct?
 
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What you have is essentially correct, although it seems bad form to use probabilities in the statement of a probability density function.

I think better is to write: Let g(z) be the PDF of Z = X + Y. If this is still the problem with Z = X + Y where X and Y are uniform on [0,1] that you've been working on in different ways, then you have this. But even if it's not you can just write it in terms of g(z).

So then:

f(z)=\left\{\begin{array}{cc}0,&amp;\mbox{ if }<br /> z &gt; 1\\ \frac{g(z)}{\int_{-\infty}^1 g(z) dz}, &amp; \mbox{ if } z \leq 1\end{array}\right
 
Last edited:


hgfalling, thank you so much for all the probability help!
 

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