Let Y be a continuous random variable with probability density function f(y) = k (2 + 3y), 1

Question

Let Y be a continuous random variable with probability density function f(y) = k (2 + 3y), 1 <y <3 and 0 elsewhere.   Find k and E(1/y)

anonymous · Answer

Let Y be a continuous random variable with probability density function f(y) = k (2 + 3y), 1 <y <3 and 0 elsewhere.   Find k and E(1/y)

amistre64 · Answer

$$\int_{1}^{3}(k2+3ky)dy=\left.2ky+\frac{3k}{2}y^2\ \right|_{1}^{3}$$

$$6k+\frac{27}{2}k-2k-\frac{3}{2}k=1$$
$$4k+\frac{24}{2}k=1$$
$$k=\frac{2}{32}=\frac{1}{16}$$
if i see it right

amistre64 · Answer

http://www.wolframalpha.com/input/?i=integrate+%281%2F16+%282%2B3y%29%29+from+1+to+3 i think its good

amistre64 · Answer

i cant tell what E(1/y) is spose to mean tho.  Maybe expected value of 1/y?

anonymous · Answer

yea i figured out the K value and you are right but didn't really found how to solve for E.... but i did (1/y)f(y) integrated from 1 to 3