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ProbablyDumd
Questions (5)
When calculating the convolution of two PDFs, one must be careful to use the appropriate limits of integration. Suppose that X
2 answers
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The random variables X and Y are independent and have the PMFs shown in this diagram;
PX(x) 1, 1/6 2, 3/6 3, 2/6 PY(y) 2, 2/6 3,
2 answers
146 views
The random variable X has a PDF of the form:
fX(x) = {1/x^2, for x >= 1, 0, otherwise.} Let Y = X^2. For Y >= 1, the PDF of Y
0 answers
151 views
Suppose that X is a continuous random variable and that Y = X^4.
Then, for y >= 0, we have: fY(y) = ay^b fX(-cy^d) + ay^b
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The random variable Q is uniform on [0, 1]. Conditioned on Q = q, the random variable X is Bernoulli with parameter q.
a) The
1 answer
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Answers (1)
The random variable X has a PDF of the form: fX(x) = {1/x^2, for x >= 1, 0, otherwise.} Let Y = X^2. For Y >= 1, the PDF of Y takes the form fY(y) = a/y^b. Find the values of a and b. a = b =
