$X_{1},...,X_{n}$ ~ $Ber(p)$ what can I say about $\exp(\lambda X_{1}),...,\exp(\lambda X_{n})$

Question

Let $X_{1},...,X_{n}$ ~ $Ber(p)$ be independent, and $\lambda > 0$. What can I say about $\exp(\lambda X_{1}),...,\exp(\lambda X_{n})$ with respect to $\mathbb E[\exp(\lambda X_{i})]$?

Well because $f_{\lambda }(x):=e^{\lambda x}$ is a continuous random variable, $(f_{\lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.

I assume $\exp(\lambda X_{i})$ will be bernoulli distributed but I fail to find a way to determine the Bernoulli coefficients, and how to prove these. Any ideas?

Answer 1

Well because $f_{\lambda }(x):=e^{\lambda x}$ is a continuous random variable, $(f_{\lambda }(X_{i}))_{i=1,...,n}$ are independent and identically distributed.

This is not true. Since $X_k$ are all discrete, then so is $\exp(\lambda X_k)$. Using the rule of discrete expectation, we can say that $$E(\exp(\lambda X)) = \sum_{x \in \lbrace 0,1 \rbrace } \Pr(X=x) \exp(\lambda x) = \Pr(X=0)\exp(\lambda (0)) + \Pr(X=1)\exp(\lambda (1))$$ Since $\Pr(X=0) = 1-p$ and $\Pr(X=1) = p$, we get $$E(\exp(\lambda X)) = 1-p+ p\exp(\lambda )$$

Answer 2

Guide:

If $X_i=0$, we have $\exp(\lambda X_i)=1$, this happens with probability $1-p$.

What happens to $\exp(\lambda X_i)$ if $X_i=1$? what is the corresponding probability.

Now use the definition of expectation to compute the desired quantity.