Suppose that a random variable X has the Bernoulli distribution with parameter p = 0.7. (See Definition 3.1.5.) Sketch the c.d.f. of X.
My attempt: So I have that the formula for the c.d.f. of a rv X is $F(x)=Pr(X\le x)$ for $-\infty<x<\infty$
Then, obviously:
$$F(x=0)=Pr(X\le 0)=0$$
$$F(x=1)=Pr(X\le 1)=1$$
Then using the fact that it has a bernoulli distribution with parameter p=0.07, I think this means that $P(X=1)=0.7$ I have an equation that says: $P(X=x)=F(x)-F(x^{-})$ so
$$P(X=x)=F(x)-F(x^{-})$$ $$P(X=x)=Pr(X \le x)-Pr(X<x)$$ $$P(X=1)=Pr(X \le 1)-Pr(X<1)$$ $$0.7=1-Pr(X<1)$$ $$Pr(X<1)=0.3$$
So I think the c.d.f. would be 0 before x=0, then jump to 0.3, and then at x=1 it would jump up to 1.
Is this correct?
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$\begingroup$A Bernouli random variables can realise just two discrete values: success $1$ and failure $0$. With a success parameter of $0.7$, this means the piecewise function is:
$$\mathsf P(X=x) =\begin{cases}0.3 & : x=0 \\ 0.7 & : x=1 \\ 0 & : \text{otherwise}\end{cases}$$
You wish to find the CDF formula, $\mathsf P(X\leqslant x)$.
So I think the c.d.f. would be 0 before x=0, then jump to 0.3, and then at x=1 it would jump up to 1.
Yes. The function is continuous except at two step discontinuities: $$\mathsf P(X\leq x) =\begin{cases} 0 & : x<0 \\ 0.3 & : 0\leqslant x < 1 \\ 1 & : 1\leqslant x\end{cases}$$
If you plot it, it will look like a staircase.
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