Question #18755

For each of the following, find the constant c so that p(x) satisfies the condition of being a probability density function of a random variable X:
I. p(x) = c(2/3)^x, x ∈ N
II. p(x) = cx, x ∈ {1,2,3,4,5,6}

Expert's answer

Question 18748 Hi, the topic is Probability Density Function of a random variable.

For each of the following, find the constant cc so that p(x)p(x) satisfies the condition of being a probability density function of a random variable XX: I. p(x)=c(2/3)xp(x) = c(2/3)^x, xNx \in N II. p(x)=cxp(x) = cx, x{1,2,3,4,5,6}x \in \{1, 2, 3, 4, 5, 6\}

I've been figuring this for couple of days. Solution. In fact, p(x)p(x) is rarely called probability density function. It is just distribution of a r.v., p(x)=P(X=x)p(x) = P(X = x).

We are to verify two conditions: 1) p(x)0p(x) \geq 0 and xNp(x)=1\sum_{x \in N} p(x) = 1. So,

a) c>0c > 0 and xNp(x)=cxN(2/3)x=c2/31/3=2c=1\sum_{x \in N} p(x) = c \sum_{x \in N} (2/3)^x = c \frac{2/3}{1/3} = 2c = 1, thus c=1/2c = 1/2. Remark: this is the case when 0N(20 \notin N(2 definitions of natural numbers are used in mathematics.), if 0N0 \in N, then xNp(x)=cxN(2/3)x=3c\sum_{x \in N} p(x) = c \sum_{x \in N} (2/3)^x = 3c, thus c=1/3c = 1/3.

b) x{1,2,3,4,5,6}cx=c672=21c\sum_{x\in \{1,2,3,4,5,6\}}cx = c\frac{6\cdot 7}{2} = 21c, thus c=1/21c = 1 / 21

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