Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. Let $X \sim U(0,1)$

ask how to calculate $E(Y)$, $E(Z)$, $E(W)$

share|cite|improve this question
Hint: $$E[g(X)] = \int_{-\infty}^{\infty} g(x)f_X(x)\mathrm dx$$ where you have been told what $f_X(x)$ is. Sketch the function $g(x) = \max(x,0.5)$ and the function $f_X(x)$, and then the function $g(x)f_X(x)$. Compute the integral above to get $E[Y] = E[\max(X,0.5)]$. Lather, rinse, repeat for the functions $\max(X-0.5,0)$ and $\max(0.5-X,0)$. – Dilip Sarwate Mar 2 '12 at 12:19
@Dilip: +1 for Lather, rinse, repeat. – Did Mar 2 '12 at 12:28
@Dilip Given that my probability exam just was over a few hours back, if not for your comment, I am a goner. I did not realise I had this tool at disposal! – user21436 Mar 2 '12 at 13:01
@KannappanSampath I call this tool LOTUS (an initialism of Law of the Unconscious Statistician) which helps jog my memory just when I need it most.... – Dilip Sarwate Mar 2 '12 at 13:10
@DilipSarwate Now I think I'll remember this rule better. Once again, Thank you for telling me! – user21436 Mar 2 '12 at 13:12

Written to elaborate the already explanatory comment of Dilip Sawarte

Let $X$ be a random variable uniformly distributed on $(0,1)$. This means that $$g_X(x)=1 ~~\text{for}~~ x \in (0,1)$$

We are interested in the expectation of the random variable, $Y=\max\left(X,\dfrac{1}{2}\right)$.

Now, note that $$\begin{align}\mathbb E(f(X))&=\int_{-\infty}^\infty f(x)g_X(x) \rm{d}x\\&=\int_{0}^1f(x)\mathrm dx\\&=\int_0^{\frac 1 2}\dfrac{1}{2}\mathrm dx+\int_{\frac 1 2}^1x~~\rm dx\\&=\dfrac 1 4+\dfrac 1 2-\dfrac 1 8\\&=\dfrac 5 8\end{align}$$

Similarly other integrals can be evaluated.

I'll leave only the answers in case you needed to check:

For (b) $\dfrac{1}{8}$

For (c) $\dfrac{1}{8}$

As Dilip Sawarte points out, some graphs you'll find useful are:

for (a):

$\hspace{1 in}$ Graph 1

for (b):

$\hspace{1 in}$ enter image description here

for (c):

$\hspace{1 in}$ Graph 3

Note that the area of the shaded region is the expectation you're in need of!

share|cite|improve this answer
Yes, true. I'll add that in. But, given that the OP is doing Continuous random variable, he should know integration as well! – user21436 Mar 2 '12 at 13:16
@Dilip I added the graphs. Thank you for the pointer. – user21436 Mar 2 '12 at 15:05
@Dilip Thanks for the pointer. GeoGebra does not allow the change AFAIK, so I should probably change the previous terminology. – user21436 Mar 2 '12 at 15:43

Consider using the formula below with $T$ being the statement that $X>0.5$,

$$ \mathbb{E}(X)=P(T)\;\mathbb{E}(X \;|\; T)+P(\text{not } T)\; \mathbb{E}(X \;|\; \text{not }T).$$

For example, for the first case we have

$$\mathbb{E}(Y)=(1/2)\; (3/4)+(1/2)\; (1/2)=5/8.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.