The continuous random variable X has probability density function fX(x) =[k(x + 3); 0

Question

The continuous random variable X has probability density function
fX(x) =[k(x + 3); 0 <=x <= 1, 0; otherwise]:
(i) Sketch the probability density function and hence, or otherwise, find the value of the
constant k
(ii) Derive E(X^t) where t > -1 and hence find the expected value of X.

anonymous · Answer

Found k to be 2/7...is that correct. But how to you do the second part?

tkhunny · Answer

$\int_{0}^{1}\dfrac{2}{7}(x+3)\cdot x^{t}\;dt$

anonymous · Answer

whats the significance of the t>1?

tkhunny · Answer

$x^{0}$ wouldn't be interesting.
$x^{1}$ would be a different matter entirely.

I think it's just a defensive restriction so we don't work the wrong problem.

tkhunny · Answer

Better thought, though.

$E[X^{0}] = 1$
$E[X^{1}] = Mean$
$E[X^{2}] = Second Moment$ Which is used to calculate Variance
$E[X^{3}] = Third Moment$ So, examining X^t, we are examining the moments of the distribution.

anonymous · Answer

so when the question says find expected value of X, I would be plugging t=1 into the derived E(X^t)?

tkhunny · Answer

That would work.  I am a little nervous that the defintion for t does not include that it is an INTEGER greater than -1.

anonymous · Answer

Yeah that is a valid point.