Find the mean and the correlation function of $Y_t=X_t-tX_1$ if $X_t$ is a Poisson process

Question

Let $\{X_t, t\geq0\}$ be a Poisson process with rate $\lambda$ and $Y_t=X_t-tX_1$, $t\in[0,1]$. Find the mean and the correlation function of $Y_t$.

$X_t$ is a Poisson process so $Pr\{X_t=k\}=\frac{(\lambda t)^k}{k!}e^{-\lambda t}$, $E[X_t]=\lambda t$. I assume for $X_1$ we will have $Pr\{X_1=k\}=\frac{(\lambda )^k}{k!}e^{-\lambda }$, $E[X_1]=\lambda$.
Now for $E[Y_t]=E[X_t-tX_1]=E[X_t]-E[tX_1]$, $t\in[0,1]$. I don't know how to calculate $E[tX_1]$ because $t$ is not a constant, but it's not a random variable either because we are not given a distribution. Should it be something like: $$E[tX_t]=\int_{0}^{1}\sum_{k=0}^{\infty}tk\frac{(\lambda t)^k}{k!}e^{-\lambda t}dt$$
For the correlation function:
$$C(Y_{t_1},Y_{t_2})=E[Y_{t_1}Y_{t_2}]-E[Y_{t_1}]E[Y_{t_2}]=$$$$=E[(X_{t_1}-t_1X_1)(X_{t_2}-t_2X_1)]-E[X_{t_1}-t_1X_1]E[X_{t_2}-t_2X_1]=$$$$=E[X_{t_1}X_{t_2}]-E[t_2X_1X_{t_1}]-E[t_1X_1X_{t_2}]+E[t_1t_2X_1^2]-E[X_{t_1}]E[X_{t_2}]+$$$$E[X_{t_1}]E[t_2X_1]+E[t_1X_1]E[X_{t_2}]-E[t_1X_1]E[t_2X_1]$$
I know that $E[X_{t_1}X_{t_2}]=\lambda t_2$ for $t_1\geq t_2$ using independence of increments, but I don't know how to calculate the other expectations.

Answer 1

$EtX_1=tEX_1$. As long as $t$ is not a random variable you can always pull it out of the expectation. Hence $EY_t=t\lambda-t\lambda=0$. Similarly $E(t_1X_{t_1})X_{t_2}=t_1 EX_{t_1}X_{t_2}$ and so on.