Consider the random variables X

x 1 | 2 | 3 P(x) 0.3 |0.5 |0.2

Find the distribution, mean µY , variance and standard deviation σY of the random variable Y = Φ(X) where (a) Φ(x) = x^3 (b) Φ(x) = 2^x.

My question is. When i will calculate the expectation value..will i have to do it based on the table above which is the distribution of x, or i have to calculate it based on the table of x^3?

  • $\begingroup$ What is the distribution of $X$? $\endgroup$ – Jonas Dec 28 '18 at 14:03
  • $\begingroup$ i have added it $\endgroup$ – Anastasia Kyriakou Dec 28 '18 at 14:07


Apply: $$\mathbb Ef(X)=\sum_x f(x)P(X=x)$$

In your case just a sum of $3$ terms, so calculators are not needed.

This in order to find $\mathbb E\Phi(X)$ and $\mathbb E\Phi(X)^2$.

Then you can also find variance and standard deviation.

  • $\begingroup$ Yes but when i will calculate the expectation value, will i do : 1*0.3+2*0.5+3*0.2 or i will do 1*0.3+4*0.5+9*0.2 because it says Φ(x) = x^3 ? $\endgroup$ – Anastasia Kyriakou Dec 28 '18 at 14:14
  • $\begingroup$ The first calculation in your comment is a calculation of $\mathbb EX=\sum_xxP(X=x)$ and the second is a calculation of $\mathbb EX^2=\sum_xx^2P(X=x)$. Since $\Phi(x)=x^3$ you should go for $\mathbb E\Phi(X)=\mathbb EX^3=\sum_xx^3P(X=x)$. $\endgroup$ – drhab Dec 28 '18 at 14:17
  • $\begingroup$ Okay and if i use only the table of x and x^2 of course to find the variance, where will i use the Φ(x) = x^3 ? $\endgroup$ – Anastasia Kyriakou Dec 28 '18 at 14:19
  • $\begingroup$ $\mathsf{Var}(\Phi(X))=\mathbb E\Phi(X)^2-(\mathbb E\Phi(X))^2$. You already found $\mathbb E\Phi(X)$ so it remains to find $\mathbb E\Phi(X)^2=\mathbb EX^6$.This with the same method. $\endgroup$ – drhab Dec 28 '18 at 14:21
  • $\begingroup$ yes i know how to calculate.but why the question says for Φ(x) = x^3? what does change? $\endgroup$ – Anastasia Kyriakou Dec 28 '18 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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