0

I'm revising probability questions after taking a break from that area for several years. I've put together a simplified version of my question and attached the probabilities as a table if that is easier to read. My question is, if you're given a distribution with negative values, do you still find the expected value and variance using the standard formula that I've put together below? I can't recall ever having seen an example with negative values before, so I was wasn't sure if it was a trick question. Also, does anyone know how to enter problems like this into Wolfram Alpha, so that I can check I got the right expected value and variance as I go through my revision problems? Expected value is usually straight forward enough, but variance calculations are so easy to mess up. Thanks!

E(X) = -1*0.4 + 1*0.35 + 3*0.25 = 0.7

enter image description here

Only_me
  • 3

1 Answers1

0

When in doubt, go for the definition.

A random varible is a function $X:\Omega \rightarrow \mathbb R$.

A continous random varible has a density function $f_X(x)$ such that $P(a\leq X \leq b)=\int_{a}^{b}f_X(x)dx$. We demand $\int_{-\infty}^{\infty}f_X(x)dx=1$.

Expectation of $X$ is defined as $\Bbb E[X]=\int_{-\infty}^{\infty}xf_X(x)dx$.

Variance of $X$ is defined as $Var(X)=\Bbb E[(X-\Bbb E[X])^2]$.

If you meet these definitions, you can do whatever you want.

Eminem
  • 651
  • 4
  • 12