# Showing $E(X^n) = n!a$ using induction. [closed]

Let $$X$$ be a random variable with $$a = E(X)$$, and suppose $$X$$ has the same distribution as $$U(Y + Z)$$, where

$$U$$ is uniformly distributed over $$(0, 1)$$, $$U, Y, Z$$ are independent and $$X, Y, Z$$ have the same distribution. How can we prove using induction that $$E(X^n) = n!a^n$$ for $$n = 1, 2,...$$.

Any hint would be appreciated.

• Expand $X^n=(U(Y+Z))^n$ inside the expectation and use independence, remember $\mathbb{E}U^n=\frac1{n+1}$. Dec 11, 2020 at 0:35
• @user10354138 I just added my attempt to your reply. Let me know if you can improve further. Dec 11, 2020 at 0:52
• So, in the statement is $n!\cdot a^n$? Dec 11, 2020 at 1:05
• @Phicar assuming statement to be true for $k=n$ we get $n!a^n=E(X^n)=E(U^n)E((Y+Z)^{n})=\frac{E((Y+Z)^{n})}{n}$ which imples that $E((Y+Z)^{n})=n.n!a^n$. If we have this then we would want to show the statement for $k=n+1$ that $E((Y+Z)^{n})=(n+1).(n+1)!a^{n+1}$ Dec 11, 2020 at 1:12
• @gunnaguy I was just pointing out that the statement is wrong in the question. it should be $a^n$ Dec 11, 2020 at 1:15

Use Strong induction (that means, suppose it holds for every $$k). Notice that $$E(U^{n+1}(Y+Z)^{n+1})=E(U^{n+1})E\left(\sum _{k=0}^{n+1}\binom{n+1}{k}Y^kZ^{n+1-k}\right )=\frac{1}{n+2}\sum _{k=0}^{n+1}\binom{n+1}{k}E(Y^k)E(Z^{n-k+1})=\frac{1}{n+2}\left (E(Z^{n+1})+E(Y^{n+1})+\underbrace{\sum_{k=1}^{n}\binom{n+1}{k}E(Y^k)E(Z^{n-k+1})}_{\text{induction here}}\right )$$ $$E(X^{n+1})=\frac{1}{n+2}\left (2E(X^{n+1})\sum_{k=1}^{n}\binom{n+1}{k}k!a^k\cdot (n-k+1)!a^{n-k+1}\right )=\frac{2E(X^{n+1})+a^{n+1}(n+1)!\cdot n}{n+2},$$ solving for $$E(X^{n+1})$$ you get $$E(X^{n+1})=(n+1)!a^{n+1}.$$

One should do a strong induction:

The base case ($$n=1$$) is essentially the definition of $$a$$.

Take $$n \geq 1$$ and assume inductively that, for all $$k = 1,\ldots,n$$, we have

$$\mathbb E (X^k) = k!a^k.$$

Then

\newcommand{\EE}{\mathbb E} \begin{alignat}4 \EE (X^{n+1}) &= \EE\big(U^{n+1}(Y+Z)^{n+1}\big) &\qquad& \\ &= \EE (U^{n+1})\EE\big((Y+Z)^{n+1}\big) &&\text{independence} \\ &= \frac1{n+2}\ \ \ \ \EE\big((Y+Z)^{n+1}\big) &&\text{hint in comments} \\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(Y^kZ^{n+1-k}) &&\text{linearity of expect.} \\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(Y^k)\EE(Z^{n+1-k}) &&\text{independence again} \\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(X^k)\EE(X^{n+1-k}) &&\text{X,Y,Z ident. distributed} \\ &= \frac2{n+2}\EE(X^{n+1})+\frac1{n+2}\sum_{k=1}^{n}\binom{n+1}k\EE(X^k)\EE(X^{n+1-k}) \\ &= \frac2{n+2}\EE(X^{n+1})+\frac1{n+2}\sum_{k=1}^{n}\frac{(n+1)!}{k!(n+1-k)!}(k!a^k)\big((n+1-k)!a^{n-1+k}\big) && \text{induct. hypothesis} \\ &= \frac2{n+2}\EE(X^{n+1}) + n!\cdot \frac n{n+2} a^{n+1} && \text{mass cancellation} \end{alignat}

and this last equality rearranges to

$$\frac{n}{n+2} \EE(X^{n+q})= \frac n {n+2}(n+1)!a^{n+1},$$

i.e. $$\EE(X^{n+1}) = (n+1)!a^{n+1},$$

completing the inductive step, and the proof.

Token remark: this implies the moment generating function no end!

$$M_X:\lambda ↦ \EE (e^{\lambda X}) = \sum_{k=0}^∞ \frac{k!a^k\lambda^k}{k!}=\cdots=\frac 1{1-a\lambda}.$$

Up to choosing $$a$$, this fixes the distribution of $$X$$. Does anyone recognise if it corresponds to something named?