wrong solution for double integral of piecewise function?

Discussion in 'Mathematica' started by Tom Roche, Feb 28, 2009.

  1. Tom Roche

    Tom Roche Guest

    wrong solution for double integral of piecewise function

    I've got a function

    f[\[Chi]_, \[Psi]_] =
    {Piecewise[{{k, {a <= \[Chi] <= b, a <= \[Psi] <= b}}}, 0]}

    I'm attempting to solve for k such that f becomes a probability
    density function by applying the normalization constraint, i.e.
    solving for k such that the double (indefinite) integral of f equals
    1. I can do this by hand pretty easily, integrating first/inside WRT
    psi and second/outside WRT chi, and I get

    (1) k = 1/((b-a)^2)

    I don't have the world's greatest calculus chops, but that looks
    correct to me. (Am I missing something?) However, when I use
    Mathematica to

    Solve[
    First[
    Integrate[
    Integrate[
    f[\[Chi], \[Psi]], {\[Psi], -\[Infinity], \[Infinity]}
    ],
    {\[Chi], -\[Infinity], \[Infinity]}
    ]
    ] == 1, k
    ]

    I get

    (2) {{k -> -(1/((a - b)^2 (-1 + UnitStep[a - b])))}}

    which seems wrong to me, though I'll admit I don't know what
    "UnitStep" means. So I'm wondering

    * does (1) = (2)? or

    * have I made a syntax error? or

    * is this just really hard to solve symbolically? If so, is there a
    better way to setup this function and its solution?
     
    Tom Roche, Feb 28, 2009
    #1
    1. Advertisements

  2. Tom Roche

    Tom Roche Guest

    Tom Roche Feb 28, 6:43 am
    Bob Hanlon
    and he points out that the proper way to do this is by wrapping an
    Assuming around the procedure above:

    Assuming[{b > a},
    Solve[
    First[
    Integrate[
    Integrate[
    f[\[Chi], \[Psi]], {\[Psi], -\[Infinity], \[Infinity]}
    ],
    {\[Chi], -\[Infinity], \[Infinity]}
    ]
    ] == 1, k
    ]
    ]

    produces

    {{k -> 1/(a - b)^2}}

    Thanks!
     
    Tom Roche, Mar 1, 2009
    #2
    1. Advertisements

  3. Tom,

    Why didn't you look up UnitStep? One press on F1 (PC) and you're done.
    It shows you that the answers are the same if a<b. I guess you have
    made the same assumption in your calculation.

    Cheers -- Sjoerd
     
    Sjoerd C. de Vries, Mar 1, 2009
    #3
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.