A nonnegative integer-valued random variable X has a CDF of the form F(x)=1-(1/5)^x for x=1,2,.....and zero otherwise. Find P(X is odd)?

Question

A nonnegative integer-valued random variable X has a CDF of the form F(x)=1-(1/5)^x
for x=1,2,.....and zero otherwise.
Find P(X is odd)?

could you please help me with that?

Answer 1

Richard is on the right track.  if P(x) = 4/5^x , then F(x+1) = F(x) + P(x+1) = 1 - (1/5)^x + 4/5^(x+1) =

1 - (1/5)^x + (5-1)/5^(x+1) = 1 - (1/5)^x + 5/5^(x+1)  - 1/5^(x+1)  = 1 - (1/5)^x + 1/5^x - 1/5^(x+1) = 1 -  1/5^(x+1)  

This is what we were trying to show (that Richard's expression for P(x) was correct.

Once we know this, P(x) = 4/5^x, then P(X is odd) is P(1) + P(3) + P(5) ...

Our first term is 4/5¹ = 4/5.  The ratio between successive odd terms 4/5^2x+1 and 4/5 ^2x+3 is 1/5² = 1/25

 

The sum of an infinite geometric sum with ratio < 1 is a/(1-r), where a is the first term and r is the ratio.

 

4/5 / (1 - 1/25) = 4/5/(24/25) = 5/6

 

A tricky way to get this result:  each odd is 5 times the next even; thus, the odds are 5 times the evens, and they sum to 1. 5/(5+1) * 1 = 5/6

 

Note: the definition of the CDF is actually wrong.  It is not 0 at the non-integers, as CDFs are monotone increasing.  Rather, it equals the value at the previous integer.

Answer 2

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Answer 3

The easiest approach is to  work out the first few values of P(x) and then look for a pattern.

 

F(1) =  1 - 1/5 =  4/5    Since there is just one term in the sum of P's at F(1)  it can be concluded that  P(1) = 4/5

 

F(2) = 1 - 1/25 = 24/25.    F(2) is  P(1) + P(2)   =  4/5 + P(2)  so   P(2) = 4/25

 

Similar logic with F(3)  leads to P(3) = 4/125.    

 

The pattern  P(x) =  4 / 5^x   can be guessed and verified because the sum of  P(x) is a geometric series.