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

Question

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.

Answer 1

Use Strong induction (that means, suppose it holds for every $k<n+1$). 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 $E(X^{n+1})=(n+1)!a^{n+1}.$

Answer 2

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.

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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?