Probability with a recursive definition

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Homework Statement

Suppose that
Pr(X = 0) = Pr(X = 1), Pr(X = k + 1) = (1/k)Pr(X = k), k = 1,2,3,···
Find Pr(0).

Homework Equations

The Attempt at a Solution

Ok I started with k = 1 and went to k = 5. The pattern I noticed is

For k=n we have p(X=n+1) = (1/2)(1/3)(1/4)...(1/n)p(X=1) = (1/n!)p(X=1)

Let k go to inf

We have Ʃ (1/n!) p(X=1)


The above summation equal 1 by probability axiom.

We get p(X=1) = Ʃ 1/n!

It is known that e^x = Ʃ xⁿ/n! Where 0≤n<∞

In this case we are starting from 1 so we must reindex to get
e^x = 1 + Ʃ xⁿ/n!

We then have

P(X=1) = 1/(Ʃ 1/n!) = 1/e-1

Since p(X=0) = p(X=1) is given we end up with

p(X=0) = 1/e-1

But the answer is 1/e+1. I sense I am close but I am messing up something with the reindexing maybe?

 

[Dick]

Homework Statement

Suppose that
Pr(X = 0) = Pr(X = 1), Pr(X = k + 1) = (1/k)Pr(X = k), k = 1,2,3,···
Find Pr(0).

Homework Equations

The Attempt at a Solution

Ok I started with k = 1 and went to k = 5. The pattern I noticed is

For k=n we have p(X=n+1) = (1/2)(1/3)(1/4)...(1/n)p(X=1) = (1/n!)p(X=1)

Let k go to inf

We have Ʃ (1/n!) p(X=1)


The above summation equal 1 by probability axiom.

We get p(X=1) = Ʃ 1/n!

It is known that e^x = Ʃ xⁿ/n! Where 0≤n<∞

In this case we are starting from 1 so we must reindex to get
e^x = 1 + Ʃ xⁿ/n!

We then have

P(X=1) = 1/(Ʃ 1/n!) = 1/e-1

Since p(X=0) = p(X=1) is given we end up with

p(X=0) = 1/e-1

But the answer is 1/e+1. I sense I am close but I am messing up something with the reindexing maybe?

You summed P(X=1)+P(X=2)+P(X=3)+... correctly. It's (Ʃ (1/n!)) p(X=1)=e*P(X=1). Then stuff starts going haywire. The axiom of probability tells you P(X=0)+P(X=1)+P(X=2)+...=1. So P(X=0)+e*p(X=1)=1. Take it from there.