I am new to statistics and I happen to came across this property of MGF:

Let $X$ and $Y$ be independent random variables. Let $Z$ be equal to $X$, with probability $p$, and equal to $Y$, with probability $1 − p$. Then,

$$M_Z(s)= p M_X(s) + (1 − p) M_Y(s).$$

The proof is given that

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]= p M_X (s) + (1 − p)M_Y (s)$$

But I do not understand, can someone show me a full proof as in showing the conditioning on the random choice between X and Y, as in why the following holds

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]$$

Thanks very much.

I am new to statistics and I happen to came across this property of MGF:

Let $X$ and $Y$ be independent random variables. Let $Z$ be equal to $X$, with probability $p$, and equal to $Y$, with probability $1 − p$. Then, $$M_Z(s)= p M_X(s) + (1 − p) M_Y(s).$$

The proof is given that

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]= p M_X (s) + (1 − p)M_Y (s)$$

But I do not understand, can someone show me a full proof as in showing the conditioning on the random choice between X and Y, as in why the following holds

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]$$

Thanks very much.

I am new to stats and I happen to came across this property of MGF

Let X and Y be independent random variables. Let Z be equal to X, with probability p, and equal to Y , with probability 1 − p. Then,

MZ(s)= pMX(s) + (1 − p)MY(s).

The proof is given that

MZ(s)= E[e^(sZ)]= pE[e^(sX)] + (1 − p)E[e^(sY)]= pMX(s) + (1 − p)MY (s).

But I do not understand, can someone show me a full proof as in showing the conditioning on the random choice between X and Y as in why

E[e^(sZ)]= pE[e^(sX)] + (1 − p)E[e^(sY)] ??

Thanks very much

I am new to statistics and I happen to came across this property of MGF:

Let $X$ and $Y$ be independent random variables. Let $Z$ be equal to $X$, with probability $p$, and equal to $Y$, with probability $1 − p$. Then,

$$M_Z(s)= p M_X(s) + (1 − p) M_Y(s).$$

The proof is given that

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]= p M_X (s) + (1 − p)M_Y (s)$$

But I do not understand, can someone show me a full proof as in showing the conditioning on the random choice between X and Y, as in why the following holds

$$M_Z(s)= E[e^{s Z}]= p E[e^{s X}] + (1 − p)E[e^{s Y}]$$

Thanks very much.