Question

Let $X$be such that$P\{X=1\}=p=1-P\{X=-1\}$

Find $c\ne 1$such that $E\left[c^X\right]=1$.

Short answer

The two possible values of $c$ are $p$ and $\frac{1}{p}-1$.

Step-by-step solution

Given information

Given in the question that,

$P\{X=1\}=p=1-P\{X=-1\}$.

Calculation

Observe that random variable $X$assumes only two values with positive probability. We have that

$P(X=1)=p$

$P(X=-1)=1-p$

Using the theorem about the expectation of a function of a random variable,

We have that

$1=E\left(c^X\right)=c\times p+\frac{1}{c}(1-p)$

Multiply both sides with $c$, we end up with an equation

$p{c}^2-c+(1-p)=0$.

Solution

Two solutions are,

$c_{1,2}=\frac{1\pm \sqrt{1-4p(1-p)}}{2p}$

Observe that the term under the root is equal to $(2p-1{)}^2$, so we have

$c_{1,2}=\frac{1\pm |2p-1|}{2p}$

Therefore, we get that we have two answers

$c_1=1,c_2=\frac{1-p}{p}$.

Final answer

There exist two possible values of $c$. They are $p$ and $\frac{1}{p}-1$.