Finding the expectation of $E[4^{Y}]$ if Y is a geometric random variable

Question

How would you find the expectation of $E[4^{Y}]$ if Y is a geometric random variable?

Answer 1

I preassume that $Y$ denotes the number of trials needed (not failures).

Denote $p:=P(Y=1)$ (chance on success).

Go for finding: $$\mathbb E[4^Y]=\sum_{n=1}^{\infty}4^nP(Y=n)=\sum_{n=1}^{\infty}4^n(1-p)^{n-1}p=4p\sum_{n=0}^{\infty}\left(4(1-p)\right)^n$$

Discern the cases $4(1-p)<1$ and $4(1-p)\geq1$, or equivalently $p>\frac34$ and $p\leq\frac34$.

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Other route:

$$\mathbb{E}[4^{Y}]=\mathbb{E}\left[4^{Y}\mid Y=1\right]P\left(Y=1\right)+\mathbb{E}\left[4^{Y}\mid Y\neq1\right]P\left(Y\neq1\right)=4p+\mathbb{E}[4^{Y}]4(1-p)$$This equality enables you to find $\mathbb{E}[4^{Y}]$.

Caution: $\mathbb E[4^{Y}]$ cannot take negative value and can take (depends on $p$) value $+\infty$.

Answer 2

Let $X = 4^Y$. Then $P(X = 4^n) = P(Y = n)$, and you know $P(Y = n)$. Now compute $E[X]$ using the series definition as usual, skipping zero summands.