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    Probability Of insurance sales




    Just a disclaimer I know I am omitting important factors such as competition, earned premium etc etc. I'm just using this exercise so I can start somewhere


    The problem that I'm trying to solve is this:
    How much impact will increasing insurance rates affect my profitability?

    Right now, let's say an insurance carrier raises rates by 10%. The incorrect way that I am thinking about it, is to increase my overall premiums by 10%, and divide losses by my new premiums amount, and that will be my impact when compared to the losses divided by the old premiums.
    The issue I'm having with this is that increasing rates will also deter business, so I'm losing more customers, more premiums, and losses, so my impact is not accurate.

    To address this, is it fair to approach it from a probability perspective? Just for example, I want to say, when the quoted amount is $200, the probability of them purchasing the policy is 20%. If I can get these probabilities, then I can know better how much new premium I will gain? And to get the amount of premium, is it calculating the expected value? (The probability * the quoted amount?)

    So to start off, I have 2 questions:
    Does the above logic make sense?
    How do I get the probability? Is it as simple as dividing the number of $200 applications by the total number of quotes?

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    Re: Probability Of insurance sales

    Does the above logic make sense?
    not to me. I read it a couple of times ... must admit I simply cannot see what you are trying to accomplish.

    I want to say, when the quoted amount is $200, the probability of them purchasing the policy is 20%.
    If this is all you want to do you can simply use a for example a logit model with independent variable being price.

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    Re: Probability Of insurance sales

    Quote Originally Posted by JesperHP View Post
    not to me. I read it a couple of times ... must admit I simply cannot see what you are trying to accomplish.


    If this is all you want to do you can simply use a for example a logit model with independent variable being price.
    Thanks. I will look into LogIt. Not too familiar with it yet, but will look into it.

    What I'm trying to accomplish is this: I want to be able to estimate how much additional premiums I can expect to gain/lose, from raising insurance rates by 10%. Suppose I quote 1000 policies per month. Not all 1000 customers will purchase because the price might be too high. By increasing rates by 10%, I can probably expect even less customers to purchase.

    So I was wondering if using probability is a way to determine how much premium I can expect to get. I,e, lets say based on historical data, my quoted amount ranges from $500 to $1000. I can figure out a way to say that quotes at a $500 price point have a 80% chance of being bound, $600 have 60%, $900 have 40% chance... etc etc. can I calculated the expected value of all that, and say that this is the amount of premiums I am expecting to gain?

    again, I'm eliminating all other factors and assuming price is th only factor

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    Re: Probability Of insurance sales

    Well under the assumption that price is the only factor and the additional assumption that customers either buy or don't - that is each customer by one unit - I would say a discrete choice model such as logit or probit with price as dependent variable is the obvious choice. Offcourse the assumptions are open to discussion - as always - but it is standard to use logit and probit to model demand. The models are simple and well known, many more refined techniques dealing with special problems exist but these models are clearly a good first choice before trying looking for more refined tools.

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    Re: Probability Of insurance sales

    Quote Originally Posted by JesperHP View Post
    Well under the assumption that price is the only factor and the additional assumption that customers either buy or don't - that is each customer by one unit - I would say a discrete choice model such as logit or probit with price as dependent variable is the obvious choice. Offcourse the assumptions are open to discussion - as always - but it is standard to use logit and probit to model demand. The models are simple and well known, many more refined techniques dealing with special problems exist but these models are clearly a good first choice before trying looking for more refined tools.
    Thanks, and as I understand it, a logit or probit will give me a single probability. I,e the probability that someone will purchase a policy that is $200. Suppose my entire quote data has premiums ranging from $200 to $1000. Would I just pick a few price points and do a logit multiple times to get each probability? I,e, do a logit for the $200 data, logit for $500 data, 750, 1000.

    And once I get that, would using the expected value be a valid way to estimate the amount of additional premium?

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    Re: Probability Of insurance sales

    Do you also have data on people being offered but ending up not buying?

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    Re: Probability Of insurance sales

    Yes, I pretty much have years of data of the quote(offer), buy/no buy, and price

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    Re: Probability Of insurance sales


    Ok then the idea is as follows .... consider a simplified version of the idea..... Every person buys y_i = 1 or don't y_i = 0.
    For simplification assume the probability of buy to be constant across individuals and denote it p.

    The expected total demand is when contacting N = 1000 customers is N times p ...
    Total demand is sum Y of individual demands Y := \sum_{i=1}^N y_i and the expectation is
    E[Y ]:= E[\sum_{i=1}^N y_i] =  \sum_{i=1}^N E[y_i] = \sum_{i=1}^N p = Np last couple of steps are valid because p is constant.

    When you estimate the logit model you get a function L(a+b \cdot price_i) = p_i that gives you the probability of buy as a function of the price.

    Inserting this in the above formula instead of the constant we have to find expected total demand:
    E[Y ]:= E[\sum_{i=1}^N y_i] =  \sum_{i=1}^N E[y_i] = \sum_{i=1}^N p_i =\sum_{i=1}^N L(a+b \cdot price_i)

    If you instead want to find the price that gives you the maximized expected revenue then the revenue is
    price_i \cdot y_i
    and the expected value of this is
    price_i \cdot E[ y_i] = price_i \cdot L(a+b \cdot price_i)

    there is a single price maximizing this expression which is intuitive since under the assumption you have no knowledge distinguishing among customers so there is no reason to offer one customer one price and another customer another.

    The whole procedure is based on the logit being a correct parametric assumption and also usual assumptions of random sampling ... these may be incorrect and
    you can check if other parametric assumption can achieve a better fit ....

    Here is some R code you may or may not understand:

    Code: 
    # Assuming the parameters are a and b are estimated
    a=0.02
    b=-0.01
    p=200  # set one price just as example
    
    # calculate L(a+b*price) this is the logit
    exp(a+b*p)/(1+exp(a+b*p))
    
    
    f=function(p)
    	{
    		p*exp(a+b*p)/(1+exp(a+b*p))
    	}
    
    price=seq(0,400,1)
    
    # Make a plot of the expected revenue for one customer ...
    # all customers are identical for all you know so maximizing revenue for one
    # does it for all:
    plot(price,f(price))
    
    # Calculate derivative of expected revenue
    g=function(p,a,b)
    	{
    		pr = exp(a+b*p)/(1+exp(a+b*p))
    		out = pr + pr*(1-pr) * b * p
    		return(out)
    
    	}
    
    # plot the derivative
    plot(price,g(price))
    
    # find the maximizing price using first order condition
    uniroot(g,a=0.02,b=-0.01,lower=0,upper=400)


    Disclaimer: I am not recommending you to use this as a business strategy, but given the assumptions the approach should be sound, unless I have made some calculation errors

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