# Product of random variable with Unif dist and its variance

1. Mar 14, 2013

### Verdict

First of, I apologize for the vague title, I didn't know how to summarize this issue.

1. The problem statement, all variables and given/known data
Suppose that the interest rate obtained in month i is a random variable
Ri with the uniform distribution on [0.01, 0.03], where R1,R2, . . . are independent.
A capital of 1 unit grows to 'The product over i, from i = 1 to n' of (1 + Ri) units in months 1, . . . , n.

a: Compute the expected capital after 12 months in an account that starts with
1 unit. (This I think I did right)

b: Compute the variance of the capital after 12 months in an account that starts
with 1 unit. (This is an issue)

c: Now suppose that a random client invests 10 units and leaves the money in the account
for N ~ Poisson(12) months, where N,R1,R2, . . . are independent random
variables.
Compute the expected capital at the time of withdrawal of this client.
(Also an issue, I think)

2. Relevant equations
The expectation value of the product of a random variable = the product of the expectation value of the random variable
Also, the expectation value of the sum of random variables = the sum of the expectation values of the random variables

And the variance is the expectation value of (the random variable)² - the mean²

Also, the expectation value for the uniform distribution is just the average of the boundaries, and of the poisson distribution it is the parameter.

3. The attempt at a solution

Alright, so for question A. What I am after is the expectation value of the product (starting from i = 1 to 12) of (1+Ri), so using the rule this is the product (from i = 1 to 12) of the expectation value of (1+Ri), which is the product of (the expectation value of 1) + (the expectation value of Ri), so it is (1+0.02)^12.

I apologize if this is rather vague, but I hope it is clear what I mean?

For question B, the variance. Well, there sadly isn't such a product rule for the variance, so I'll probably have to use the 'original' definition, somehow. However, I don't really see what to do. Do I take the product (starting from i = 1 to 12) of (1+Ri) as my random variable, and then square that, and compute it's expectation value? If so, how can I do that?

For question C, I am a bit puzzled. Seeing it first I thought of the conditional expectation value, but that doesn't work, as the Poisson distribution is discrete and the uniform distribution is continuous. So clearly something else has to be going on. However, the expectation value of 1 unit of currency after 12 months I know from A. The expectation value of poisson(12) is also just 12, so is it just the answer to A, multiplied by 10? Seems to simple.

Verdict

2. Mar 14, 2013

### Ray Vickson

$$\text{Var}\left[ \prod_{i=1}^n (1+R_i) \right] = E \left[ \prod_{i=1}^n (1+R_i) \right]^2 - \left[ E \prod_{i=1}^n (1+R_i) \right]^2,$$
and for independent $R_i$ we have
$$\prod_{i=1}^n E\,[(1+R_i)^2] - \prod_{i=1}^n [E(1+R_i)]^2.$$

For (c): for capital C, the expected value is
$$EC = P(N=0)E[C|N=0] + P(N=1) E[C|N=1] + P(N=2) E[C|N=2] + \cdots .$$

3. Mar 14, 2013

### Verdict

Hmm. I understand (and agree) with the first line, that is indeed what I tried. However, simplifiying it to what you have written down is not completely obvious for me. How exactly do you do that? As in, how do you pull out the product of the square? Thinking about it in my head, I suppose it does make sense, its just rearranging the terms. Alright, I'll do it that way, thanks!

For C, I don't really understand what you mean by that I am afraid. It's just a weighted average of all the possible values of N?

4. Mar 14, 2013

### Ray Vickson

Right. If N = 1 you need conditional expectation of C, *given N = 1*. If N = 2 you need the conditional expectation of C, *given N = 2*, etc. This is a standard probability tool called a "conditioning argument". If you have not heard of it before, now is the time to learn about it; it simplifies things tremendously. See, eg.,
http://en.wikipedia.org/wiki/Conditioning_(probability) .

Alternatively, you can use
$$P\{C \in (x,x + \Delta x) \} = P\{C \in (x,x + \Delta x) \: \& \: N = 0\} + P\{C \in (x,x + \Delta x) \: \& \: N = 1\} + P\{C \in (x,x + \Delta x) \: \& \: N = 2 \} + \cdots$$

5. Mar 14, 2013

### Verdict

Hm alright, I'll give that a read after dinner. However, when I compute b, I get that the variance is 0. This does seem rather odd does it not?
I used that 'the product from i = 1 to 12' of E[(1+Ri)²] = 'the product..' of [E(1) + 2E(Ri) + E(Ri)E(Ri)] and filling in the numbers, I get exactly zero. Is that correct?

Also for C, I do know about conditional distributions, but I don't see how that works with a continuous variable given a discrete variable?

6. Mar 14, 2013

### Ray Vickson

If I tell you that {N=1} occurs, can you tell me what is the random variable C (or, at least, its expectation)? If I tell you that {N=2} occurs, can you tell me the random variable C or its expectation? Ditto for {N = 3}, etc. That's all there is to it!

7. Mar 15, 2013

### Verdict

Alright, so I think I'm on to it then.

The pmf of 1 unit of capital given N = n, I would say is just (1.02)^n
Then, multiplying it with the chance of n being that value for all n, I get
The sum from n = 0 to infinity of e^-12 * 12^n / n! * (1.02)^n

Now, just by rewriting this a little it reduces to e^0.24
And then the final answer, of the expectation value of 10 units of capital when N has a poisson(12) distribution, is 10*e^.24 = 12.713, which is very similar to the answer of a) multiplied by 10, which gives me confidence that this is at least on the right track! :)

This is basically the rule of the lazy statistician, am I correct?

Last edited: Mar 15, 2013
8. Mar 15, 2013

### Ray Vickson

More-or-less, although I have only heard it called the "law of the unconscious statistician".

9. Mar 15, 2013

### Verdict

I suppose my book just uses uncommon notation then. Either way, thanks a bunch, that really helped!