Let X be an exponential random variable with pdf fx(x)=-*/2, x ≥0. Let Y = floor(X), where floor(a) is the largest integer less than or equal to x (i.e. it rounds a down), so floor(0.142) = 0, floor(4.9982) = 4, etc. Note that Y is a discrete random variable, since it can only take the values 0, 1, 2,..... (a) Find the pmf py (k) for Y. (Hint: If you are having trouble, maybe start by calculating py (0) = P(Y = 0), then Py (1) P(Y 1), then generalize.) = (b) Let A be the event {Y is even}. Find an expression for the conditional pmf py|A(k). Handy fact: a² = when |a| < 1.

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Let X be an exponential random variable with pdf
fx(x) = ½³/2, x≥0.
Let Y = floor(X), where floor(x) is the largest integer less than or equal to x (i.e. it rounds a
down), so floor(0.142) = 0, floor(4.9982) = 4, etc. Note that Y is a discrete random variable,
since it can only take the values 0, 1, 2,....
(a) Find the pmf py(k) for Y.
==
(Hint: If you are having trouble, maybe start by calculating py (0) = P(Y = 0), then
Py (1) = P(Y = 1), then generalize.)
(b) Let A be the event {Y is even}. Find an expression for the conditional pmf py|A(k).
Handy fact: a = when |a| < 1.
Transcribed Image Text:Let X be an exponential random variable with pdf fx(x) = ½³/2, x≥0. Let Y = floor(X), where floor(x) is the largest integer less than or equal to x (i.e. it rounds a down), so floor(0.142) = 0, floor(4.9982) = 4, etc. Note that Y is a discrete random variable, since it can only take the values 0, 1, 2,.... (a) Find the pmf py(k) for Y. == (Hint: If you are having trouble, maybe start by calculating py (0) = P(Y = 0), then Py (1) = P(Y = 1), then generalize.) (b) Let A be the event {Y is even}. Find an expression for the conditional pmf py|A(k). Handy fact: a = when |a| < 1.
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